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Mathematics I
Chapter 10
Dr. Devendra Kumar
Department of Mathematics
Birla Institute of Technology & Science, Pilani
20152016
Devendra Kumar BITS, Pilani Mathematics I
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CHAPTER 10
Infinite Series
Devendra Kumar BITS, Pilani Mathematics I
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In this Chapter, we are going to learn about Infinite
series and their convergence properties. We will be
discussing the following topics:
Sequences (Certain Theorems on Sequences)
Devendra Kumar BITS, Pilani Mathematics I
-
In this Chapter, we are going to learn about Infinite
series and their convergence properties. We will be
discussing the following topics:
Sequences (Certain Theorems on Sequences)
Infinite Series
Devendra Kumar BITS, Pilani Mathematics I
-
In this Chapter, we are going to learn about Infinite
series and their convergence properties. We will be
discussing the following topics:
Sequences (Certain Theorems on Sequences)
Infinite Series
Integral Test
Devendra Kumar BITS, Pilani Mathematics I
-
In this Chapter, we are going to learn about Infinite
series and their convergence properties. We will be
discussing the following topics:
Sequences (Certain Theorems on Sequences)
Infinite Series
Integral Test
Comparison Tests
Devendra Kumar BITS, Pilani Mathematics I
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Ratio and Root Tests
Devendra Kumar BITS, Pilani Mathematics I
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Ratio and Root Tests
Alternating Series
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Ratio and Root Tests
Alternating Series
Power Series
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Ratio and Root Tests
Alternating Series
Power Series
Taylor & Maclaurin Series
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Section 10.1
Sequences [Self Study]
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Sequence
A function whose domain is the set of natural
numbers ( f :NR) is called a sequence of realnumbers. We write f (n)= an, then the sequence isdenoted by {a1,a2, . . .} or by {an}
n=1 or simply by {an}.
We call an the nth term of the sequence or the value
of the sequence at n.
Devendra Kumar BITS, Pilani Mathematics I
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Examples
{n}= {1,2,3, . . . ,n, . . .}
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Examples
{n}= {1,2,3, . . . ,n, . . .}{1n
}=
{1, 1
2, 13, . . . , 1
n, . . .
}
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Examples
{n}= {1,2,3, . . . ,n, . . .}{1n
}=
{1, 1
2, 13, . . . , 1
n, . . .
}{(1)n+1
}= {1,1,1,1, . . . , (1)n+1, . . .}
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Examples
{n}= {1,2,3, . . . ,n, . . .}{1n
}=
{1, 1
2, 13, . . . , 1
n, . . .
}{(1)n+1
}= {1,1,1,1, . . . , (1)n+1, . . .}{
1 1n
}=
{0, 1
2, 23, . . . ,1 1
n, . . .
}
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Convergent Sequence
A sequence {an} is said to converge to a number L, if
for every positive number , however small, we can
find a positive integer N (depending on ) such that
|anL| < , n>N.
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If no such number L exists, the sequence is said to
diverge.
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If no such number L exists, the sequence is said to
diverge.
If {an} converges to L, then we write limnan = L or
simply an L.
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If no such number L exists, the sequence is said to
diverge.
If {an} converges to L, then we write limnan = L or
simply an L.The number L is called the limit of the sequence.
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If no such number L exists, the sequence is said to
diverge.
If {an} converges to L, then we write limnan = L or
simply an L.The number L is called the limit of the sequence.
A sequence can not converges to more than one
limit i.e., Limit of a sequence is unique.
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Graphical Representation
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Graphical Representation
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Examples
The sequence{1n
}converges to the number 0
(using Archimedean property).
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Examples
The sequence{1n
}converges to the number 0
(using Archimedean property).
The sequence {n} diverges.
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Examples
The sequence{1n
}converges to the number 0
(using Archimedean property).
The sequence {n} diverges.
The sequence{(1)nn
}converges to the number 0
(using Archimedean property).
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Examples
The sequence{1n
}converges to the number 0
(using Archimedean property).
The sequence {n} diverges.
The sequence{(1)nn
}converges to the number 0
(using Archimedean property).
The sequence{1+ 1
n
}converges to the number 1
(using Archimedean property).
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Examples
The sequence{1n
}converges to the number 0
(using Archimedean property).
The sequence {n} diverges.
The sequence{(1)nn
}converges to the number 0
(using Archimedean property).
The sequence{1+ 1
n
}converges to the number 1
(using Archimedean property).
The sequence {(1)n} diverges.Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 1.)
Let {an} and {bn} be two convergent sequences. Then
1 limkan = k liman2 lim(anbn)= liman limbn3 lim(an.bn)= liman. limbn4 lim
(anbn
)= liman
limbn, if bn , 0 for all n and limbn , 0.
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Theorem (Theorem 2. Sandwich Theorem or
Squeeze Theorem)
Let {an}, {bn}, and {cn} be sequences of real numbers.
If an bn cn for all nN and if liman = lim cn = L,then limbn = L.
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Theorem (Theorem 2. Sandwich Theorem or
Squeeze Theorem)
Let {an}, {bn}, and {cn} be sequences of real numbers.
If an bn cn for all nN and if liman = lim cn = L,then limbn = L.
Q:63 Find lim n!nn
(If exists).
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Theorem (Theorem 2. Sandwich Theorem or
Squeeze Theorem)
Let {an}, {bn}, and {cn} be sequences of real numbers.
If an bn cn for all nN and if liman = lim cn = L,then limbn = L.
Q:63 Find lim n!nn
(If exists).
Sol. 0< n!nn= n(n1)(n2)...3.2.1
n.n.n...n.n.n 1
n 0. So an
application of Sandwich theorem gives lim n!nn= 0.
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Theorem (Theorem 3. Continuous Function
Theorem)
Let {an} be a sequence of real numbers. If an L andif f (x) is a function that is continuous at x= L anddefined at all an, then f (an) f (L).
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Theorem (Theorem 3. Continuous Function
Theorem)
Let {an} be a sequence of real numbers. If an L andif f (x) is a function that is continuous at x= L anddefined at all an, then f (an) f (L).
Q:43 Find limsin(pi
2+ 1
n
)(If exists).
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Theorem (Theorem 3. Continuous Function
Theorem)
Let {an} be a sequence of real numbers. If an L andif f (x) is a function that is continuous at x= L anddefined at all an, then f (an) f (L).
Q:43 Find limsin(pi
2+ 1
n
)(If exists).
Sol. Let an = pi2 + 1n . Then
liman = lim(pi
2+ 1n
)= limpi
2+ lim 1
n= pi2.
So using continuous function theorem
limsin(pi
2+ 1
n
)= sin pi
2= 1.
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Theorem (Theorem 4.)
Suppose that f (x) is a function defined for all x n0and that {an} is a sequence of real numbers s.t.
an = f (n) for n n0, thenlimx f (x)= L limnan = L.
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Q:50 Find lim lnnln2n
.
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Q:50 Find lim lnnln2n
.
Sol.
limln x
ln2x= lim
1x
22x
, by LHospitals rule
= 1.
Hence lim lnnln2n
= 1.
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Bounded Sequence
A sequence {an} is said to be bounded from above if
there exists a number M1 such that
an M1, n.The number M1 is called an upper bound for {an}.
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Bounded Sequence
A sequence {an} is said to be bounded from above if
there exists a number M1 such that
an M1, n.The number M1 is called an upper bound for {an}.
Similarly, a sequence {an} is said to be bounded from
below if there exists a number M2 such that
an M2, n.The number M2 is called a lower bound for {an}.
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Bounded Sequence
A sequence {an} is said to be bounded if there exists a
number M such that
|an| M, n.The number M is called a bound for {an}.
Examples
The sequence{
nn+1
}is bounded below by 1
2and
bounded above by 1.
Devendra Kumar BITS, Pilani Mathematics I
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Bounded Sequence
A sequence {an} is said to be bounded if there exists a
number M such that
|an| M, n.The number M is called a bound for {an}.
Examples
The sequence{
nn+1
}is bounded below by 1
2and
bounded above by 1.
The sequence{12n
}is bounded below by 0 and
bounded above by 12.
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Nondecreasing/Nonincreasing/Monotonic
Sequence
A sequence {an} is called a nondecreasing sequence,
if an an+1 for all n. The sequence is nonincreasing,if an an+1 for all n. The sequence is monotonic if itis either nondecreasing or nonincreasing.
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Nondecreasing/Nonincreasing/Monotonic
Sequence
A sequence {an} is called a nondecreasing sequence,
if an an+1 for all n. The sequence is nonincreasing,if an an+1 for all n. The sequence is monotonic if itis either nondecreasing or nonincreasing.
Examples
The sequence{
nn+1
}is nondecreasing.
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Nondecreasing/Nonincreasing/Monotonic
Sequence
A sequence {an} is called a nondecreasing sequence,
if an an+1 for all n. The sequence is nonincreasing,if an an+1 for all n. The sequence is monotonic if itis either nondecreasing or nonincreasing.
Examples
The sequence{
nn+1
}is nondecreasing.
The sequence{12n
}is nonincreasing.
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Theorem (Theorem 6. Monotonic Sequence
Theorem)
A bounded monotonic sequence is convergent.
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Theorem (Theorem 6. Monotonic Sequence
Theorem)
A bounded monotonic sequence is convergent.
Example
The sequence{1 1
2n
}is monotonic (nondecreasing)
and bounded (as |an| 1), therefore it is convergent.
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Homework
Using following two facts:
The converse of Theorem 6 is not true.
Every convergent sequence is bounded (Theorem).
Give a counter example of a convergent sequence
which is not monotonic.
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Section 10.2
Infinite Series
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Infinite series
An infinite series is the sum of the terms of a
sequence. Thus an infinite series consists of infinite
number of real numbers separated by + sign. Thusfor a sequence {an} the series will be written as
a1+a2+a3+ +an+ or asn=1
an or simply asan. The number an is called the n
th term of the
series.
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Examples
1+ 12+ 1
3+ + 1
n+ (Harmonic Series)
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Examples
1+ 12+ 1
3+ + 1
n+ (Harmonic Series)
11+11+ + (1)n+1+
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Examples
1+ 12+ 1
3+ + 1
n+ (Harmonic Series)
11+11+ + (1)n+1+ 1+ 1
22+ 1
32+ + 1
n2+
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Sequence of Partial Sums
Consider the seriesan. The sequence {Sn}, defined
by
Sn = a1+a2+a3+ +an, nis called the sequence of partial sums of the series.
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Sequence of Partial Sums
Consider the seriesan. The sequence {Sn}, defined
by
Sn = a1+a2+a3+ +an, nis called the sequence of partial sums of the series.
The nth term Sn of the sequence is called nth partial
sum of the series.
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Convergent Series
A seriesan is said to be convergent iff the sequence
{Sn} is convergent. If {Sn} converges to a real number
L, then the seriesan also converges to L and the
sum of the series is L.
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Convergent Series
A seriesan is said to be convergent iff the sequence
{Sn} is convergent. If {Sn} converges to a real number
L, then the seriesan also converges to L and the
sum of the series is L.
A series which is not convergent is called divergent.
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Telescoping Series 1
n(n+1)Here an = 1n(n+1) , the Sn can be explicitly evaluatedas follows:
an =1
n(n+1) =1
n 1n+1.
Thus
Sn =(1 1
2
)+
(1
2 13
)+ +
(1
n 1n+1
)= 1 1
n+1.
Thus limSn = 1 and hence 1
n(n+1) = 1.
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Q:3 Find the formula for nth partial sum and use it
to find the series sum if the series converges:
1 12+ 14 18+ + (1)n1 1
2n1+
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Q:3 Find the formula for nth partial sum and use it
to find the series sum if the series converges:
1 12+ 14 18+ + (1)n1 1
2n1+
Sol. Here Sn =1(1
(12
)n)1(12)
, therefore
limSn = 11(12)
= 23. Thus
1 12+ 14 18+ + (1)n1 1
2n1+ = 2
3.
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Remark.
Adding or deleting a finite number of terms in a
series does not alter the behavior (convergence or
divergence) of the series. However the sum of the
series will change in the case of convergent series.
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There are two well known series which will be used
quite often:
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There are two well known series which will be used
quite often:
1 The geometric series:arn1, a, 0.
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There are two well known series which will be used
quite often:
1 The geometric series:arn1, a, 0.
2 The p-series: 1
np, p R.
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Theorem
The geometric seriesarn1 with a, 0, is
(i) convergent if |r| < 1 and arn1 = a1r .
(ii) divergent if |r| 1.
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Theorem
The geometric seriesarn1 with a, 0, is
(i) convergent if |r| < 1 and arn1 = a1r .
(ii) divergent if |r| 1.Proof.Well prove the theorem by using the
definition of convergence. We have
Sn = a+ar+ar2+ . . .+arn1 =a(1 rn)1 r , r , 1.
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Theorem
The geometric seriesarn1 with a, 0, is
(i) convergent if |r| < 1 and arn1 = a1r .
(ii) divergent if |r| 1.Proof.Well prove the theorem by using the
definition of convergence. We have
Sn = a+ar+ar2+ . . .+arn1 =a(1 rn)1 r , r , 1.
(i) First we consider the case when |r| < 1, in thiscase rn 0 as n. Therefore, Sn a1r as n,and hence the GS is convergent. Also, its sum in this
case is a1r .
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(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.
Devendra Kumar BITS, Pilani Mathematics I
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-
(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r 1: We have rn as n and hence Snalso tends to or (depending on the sign ofa).
Devendra Kumar BITS, Pilani Mathematics I
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(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r 1: We have rn as n and hence Snalso tends to or (depending on the sign ofa).
For r
-
(ii) Now we consider the case when |r| 1. Again wesplit this case as |r| > 1 and |r| = 1.Let we first consider |r| > 1: Then r > 1 or r 1: We have rn as n and hence Snalso tends to or (depending on the sign ofa).
For r
-
Now consider |r| = 1: Then r = 1 or r =1.
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Now consider |r| = 1: Then r = 1 or r =1.If r = 1, then GS becomes a+a+ . . ., so thatSn = na; this tends to or as n(depending on the sign of a); and hence GS is
divergent in this case also.
Devendra Kumar BITS, Pilani Mathematics I
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Now consider |r| = 1: Then r = 1 or r =1.If r = 1, then GS becomes a+a+ . . ., so thatSn = na; this tends to or as n(depending on the sign of a); and hence GS is
divergent in this case also.
If r =1, then GS becomes aa+aa+ . . ., so that
Sn ={0, if n is even
a, if n is odd.
Since a is not zero, Sn does not tend to a unique
limiting value and hence is divergent in this case
also. Devendra Kumar BITS, Pilani Mathematics I
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Q:50n=0
(p2)n.
Devendra Kumar BITS, Pilani Mathematics I
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Q:50n=0
(p2)n.
Sol. Here r =p2> 1, therefore
n=0
(p2)n is
divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:50n=0
(p2)n.
Sol. Here r =p2> 1, therefore
n=0
(p2)n is
divergent.
Q:57 2
10n.
Devendra Kumar BITS, Pilani Mathematics I
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Q:50n=0
(p2)n.
Sol. Here r =p2> 1, therefore
n=0
(p2)n is
divergent.
Q:57 2
10n.
Sol. Here r = 110< 1, therefore 2
10nis convergent
and 210n
=210
1 110
= 29.
Devendra Kumar BITS, Pilani Mathematics I
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Q:67n=0
(epi
)n.
Devendra Kumar BITS, Pilani Mathematics I
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Q:67n=0
(epi
)n.
Sol. Here r = epi< 1, therefore n=0 ( epi)n is convergent
and n=0
( epi
)n= 11 e
pi
= pipi e .
Devendra Kumar BITS, Pilani Mathematics I
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Q:67n=0
(epi
)n.
Sol. Here r = epi< 1, therefore n=0 ( epi)n is convergent
and n=0
( epi
)n= 11 e
pi
= pipi e .
Q:68n=0
enpi
pine .
Devendra Kumar BITS, Pilani Mathematics I
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Q:67n=0
(epi
)n.
Sol. Here r = epi< 1, therefore n=0 ( epi)n is convergent
and n=0
( epi
)n= 11 e
pi
= pipi e .
Q:68n=0
enpi
pine .
Sol. Here r = epipie > 1 (How ?), therefore
n=0
enpi
pine is
divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 7. The necessary condition
for a series to be convergent)
Ifan converges, then liman = 0.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 7. The necessary condition
for a series to be convergent)
Ifan converges, then liman = 0.
Proof.We have
a1 = S1an = SnSn1 for n= 2,3,4, . . .
If the given series converges then Sn L. Therefore,liman = lim(SnSn1) LL= 0.
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Remark.
The converse of above theorem is not true
i.e., liman = 0 does not necessarily imply that theseries
an converges.
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Remark.
The converse of above theorem is not true
i.e., liman = 0 does not necessarily imply that theseries
an converges.
Example
For the Harmonic series 1
n; liman = 0 but the series
is divergent (well see later). Thus we have following
important test for the divergence.
Devendra Kumar BITS, Pilani Mathematics I
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The nth term test for divergence
If liman fails to exist or is different from zero, thenan is divergent.
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The nth term test for divergence
If liman fails to exist or is different from zero, thenan is divergent.
Q:34 or 53n=0
cosnpi.
Devendra Kumar BITS, Pilani Mathematics I
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The nth term test for divergence
If liman fails to exist or is different from zero, thenan is divergent.
Q:34 or 53n=0
cosnpi.
Sol. Here liman = limcosnpi= lim(1)n, which doesnot exist, therefore
n=0
cosnpi is divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:60(
1 1n
)n.
Devendra Kumar BITS, Pilani Mathematics I
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Q:60(
1 1n
)n.
Sol. liman = lim(1 1
n
)n = lim(1+ 1n
)n = e1 , 0,therefore
(1 1
n
)nis divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:60(
1 1n
)n.
Sol. liman = lim(1 1
n
)n = lim(1+ 1n
)n = e1 , 0,therefore
(1 1
n
)nis divergent.
Q:62 nn
n!.
Devendra Kumar BITS, Pilani Mathematics I
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Q:60(
1 1n
)n.
Sol. liman = lim(1 1
n
)n = lim(1+ 1n
)n = e1 , 0,therefore
(1 1
n
)nis divergent.
Q:62 nn
n!.
Sol. liman = lim nn
n!= (How?), therefore nn
n!is
divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:66ln
(n
2n+1).
Devendra Kumar BITS, Pilani Mathematics I
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Q:66ln
(n
2n+1).
Sol. lim n2n+1 = lim 12+ 1n =
12, therefore
liman = limln(
n2n+1
)= ln 1
2, 0, and so
ln
(n
2n+1)is
divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 8.)
Ifan = A and
bn =B are two convergent series,
then
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 8.)
Ifan = A and
bn =B are two convergent series,
then(anbn)=
an
bn = AB.
kan = kan = kA (for any real number k).
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Section 10.3
The Integral Test
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For next few classes, series will be assumed to be
series of nonnegative terms i.e., an 0 for all n. Thismeans that Sn Sn+1 for all n. That is the sequence{Sn} is nondecreasing.
Devendra Kumar BITS, Pilani Mathematics I
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For next few classes, series will be assumed to be
series of nonnegative terms i.e., an 0 for all n. Thismeans that Sn Sn+1 for all n. That is the sequence{Sn} is nondecreasing.
Corollary (Corollary of Theorem 6)
A seriesan of nonnegative terms converges iff Sn is
bounded from above.
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Theorem (Theorem 9. The Integral Test)
Letan be a series of positive terms. Let f (x) be a
positive, continuous and decreasing function for all
xN for some N; and let f (n)= an for all n. Thenthe series
n=N
an and the integralN f (x)dx both
converge or both diverge.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 9. The Integral Test)
Letan be a series of positive terms. Let f (x) be a
positive, continuous and decreasing function for all
xN for some N; and let f (n)= an for all n. Thenthe series
n=N
an and the integralN f (x)dx both
converge or both diverge.
Remark.
The sum of the series and the value of integral need
not necessarily be equal in the convergent case.
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Let we apply this test to p-series:
An Important Result
The p-series 1
npis convergent for p> 1, and
divergent for p 1.
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Let we apply this test to p-series:
An Important Result
The p-series 1
npis convergent for p> 1, and
divergent for p 1.Proof. For p 0, the series is trivially divergent (asliman , 0).
Devendra Kumar BITS, Pilani Mathematics I
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Let we apply this test to p-series:
An Important Result
The p-series 1
npis convergent for p> 1, and
divergent for p 1.Proof. For p 0, the series is trivially divergent (asliman , 0).
So we consider p> 0. Note that for p> 0 the functionf (x)= 1
xpis positive, continuous and decreasing when
x 1. So we can apply integral test:
Devendra Kumar BITS, Pilani Mathematics I
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If p> 1, then1 f (x)dx =
1
1xpdx= limb
[xp+1p+1
]b1= 1
p1.Thus, the series converges in this case.
Devendra Kumar BITS, Pilani Mathematics I
-
If p> 1, then1 f (x)dx =
1
1xpdx= limb
[xp+1p+1
]b1= 1
p1.Thus, the series converges in this case.
If 0< p 1. We split the case
Devendra Kumar BITS, Pilani Mathematics I
-
If p> 1, then1 f (x)dx =
1
1xpdx= limb
[xp+1p+1
]b1= 1
p1.Thus, the series converges in this case.
If 0< p 1. We split the caseIf p= 1, we get
1
1xdx= limb[ln x]b1. So the
series is divergent.
Devendra Kumar BITS, Pilani Mathematics I
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If p> 1, then1 f (x)dx =
1
1xpdx= limb
[xp+1p+1
]b1= 1
p1.Thus, the series converges in this case.
If 0< p 1. We split the caseIf p= 1, we get
1
1xdx= limb[ln x]b1. So the
series is divergent.
If 0< p< 1, then1 f (x)dx=
1
1xpdx= limb
[xp+1p+1
]b1. Hence the
p-series diverges in this case also.
Devendra Kumar BITS, Pilani Mathematics I
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If p> 1, then1 f (x)dx =
1
1xpdx= limb
[xp+1p+1
]b1= 1
p1.Thus, the series converges in this case.
If 0< p 1. We split the caseIf p= 1, we get
1
1xdx= limb[ln x]b1. So the
series is divergent.
If 0< p< 1, then1 f (x)dx=
1
1xpdx= limb
[xp+1p+1
]b1. Hence the
p-series diverges in this case also.
An Important Result
By the above test the Harmonic series 1
ndiverges.
Devendra Kumar BITS, Pilani Mathematics I
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Q:18 Decide whether the series converges/diverges:
lnnn2
.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18 Decide whether the series converges/diverges:
lnnn2
.
Sol. The function f (x)= ln xx2
is positive, continuous
and decreasing for all x 2 (note that these threeconditions holds true for all x> pe). Now, we have
Devendra Kumar BITS, Pilani Mathematics I
-
2
ln x
x2dx=
ln2
tet dt (on substituting ln x= t)
= limb
[(t+1)et]bln2
= 1+ ln22
.
Hence by integral test the series lnn
n2is convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:32 1
n(1+ln2n).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:32 1
n(1+ln2n).
Sol. The function f (x)= 1x(1+ln2 x) is positive,
continuous and decreasing for all x 1.1
1
x(1+ ln2 x)dx=
0
dt
1+ t2 (on substituting lnx= t)
= limb
[tan1 t]b0
= pi2.
Hence by integral test the series is convergent.
Devendra Kumar BITS, Pilani Mathematics I
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HomeworkWhich of the series converge and which
diverge? 1n.
Devendra Kumar BITS, Pilani Mathematics I
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HomeworkWhich of the series converge and which
diverge? 1n. 1
n2.
Devendra Kumar BITS, Pilani Mathematics I
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HomeworkWhich of the series converge and which
diverge? 1n. 1
n2.
Ans. Divergent, Convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
HomeworkWhich of the series converge and which
diverge? 1n. 1
n2.
Ans. Divergent, Convergent.
Homework For what p> 0, does the series2
1n(lnn)p
converge?
Devendra Kumar BITS, Pilani Mathematics I
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HomeworkWhich of the series converge and which
diverge? 1n. 1
n2.
Ans. Divergent, Convergent.
Homework For what p> 0, does the series2
1n(lnn)p
converge?
Ans. p> 1.
Devendra Kumar BITS, Pilani Mathematics I
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Section 10.4
Comparison Tests
Devendra Kumar BITS, Pilani Mathematics I
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Now, well study two comparison tests:
Devendra Kumar BITS, Pilani Mathematics I
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Now, well study two comparison tests:
The Direct Comparison Test (DCT)
Devendra Kumar BITS, Pilani Mathematics I
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Now, well study two comparison tests:
The Direct Comparison Test (DCT)
The Limit Comparison Test (LCT)
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 10. The Direct Comparison
Test)
Letan be a series of nonnegative terms. Then
(a) If an cn for all nN, wherecn is a
known convergent series; thenan
converges.
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 10. The Direct Comparison
Test)
Letan be a series of nonnegative terms. Then
(a) If an cn for all nN, wherecn is a
known convergent series; thenan
converges.
(b) If an dn for all nN, wheredn is a
known divergent series of nonnegative terms,
thenan diverges.
Devendra Kumar BITS, Pilani Mathematics I
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Example
As an application, well prove that the exponential
series 1
n!is convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Example
As an application, well prove that the exponential
series 1
n!is convergent.
Sol. Here, an = 1n! . We know that 2n1 n! for all n.Therefore, we have,
an =1
n! 12n1
= cn (say) for all n.
Moreover, 1
2n1 , being a geometric series with r =12,
is convergent. Therefore, by DCT the exponential
series is convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:18 3
n+pn
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18 3
n+pnSol. Here an = 3n+pn
3/2n
for all n. Since the series 1ndiverges, therefore by DCT
3n+pn diverges.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18 3
n+pnSol. Here an = 3n+pn
3/2n
for all n. Since the series 1ndiverges, therefore by DCT
3n+pn diverges.
Q:26 1p
n3+2
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18 3
n+pnSol. Here an = 3n+pn
3/2n
for all n. Since the series 1ndiverges, therefore by DCT
3n+pn diverges.
Q:26 1p
n3+2Sol. Here an = 1p
n3+2 1
n3/2for all n. Since the series 1
n3/2converges, therefore by DCT
1pn3+2
converges.
Devendra Kumar BITS, Pilani Mathematics I
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Q:44 (n1)!
(n+2)!.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:44 (n1)!
(n+2)!.Sol. Here an = (n1)!(n+2)! = 1(n+2)(n+1)n < 1n3 for all n. Sincethe series
1n3
converges, therefore by DCT (n1)!
(n+2)!converges.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (11. Limit Comparison Test)
Suppose an and bn are positive for all nN.(a) If lim
anbn= c (finite and non zero), then both
an andbn converge or diverge together.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (11. Limit Comparison Test)
Suppose an and bn are positive for all nN.(a) If lim
anbn= c (finite and non zero), then both
an andbn converge or diverge together.
(b) If limanbn= 0, and bn converges, then an
also converges.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (11. Limit Comparison Test)
Suppose an and bn are positive for all nN.(a) If lim
anbn= c (finite and non zero), then both
an andbn converge or diverge together.
(b) If limanbn= 0, and bn converges, then an
also converges.
(c) If limanbn=, and bn diverges, then an
also diverges.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:24
n=35n33n
n2(n2)(n2+5).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:24
n=35n33n
n2(n2)(n2+5).
Sol. Here an = 5n33n
n2(n2)(n2+5). Consider bn =n3
n5= 1
n2.
Note thatbn is convergent. Now
liman
bn= lim
5n33nn2(n2)(n2+5)
1n2
= 5.
Hencean is convergent, by LCT (a).
Devendra Kumar BITS, Pilani Mathematics I
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Q:36 n+2n
n22n
Devendra Kumar BITS, Pilani Mathematics I
-
Q:36 n+2n
n22n
Sol. Here an = n+2n
n22n. Since 2n > n for all n so consider
bn = 2n
n22n= 1
n2. Note that
bn is convergent. Now
liman
bn= lim
n+2nn22n
1n2
= lim n+2n
2n= 1.
Hencean is convergent, by LCT (a).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:28 (lnn)2
n3.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:28 (lnn)2
n3.
Sol. Here an = (lnn)2
n3. Consider bn = n
c
n3= 1
n3c , where c
is very small (close to zero). Note thatbn is
convergent. Now
liman
bn= lim
(lnn)2
n3
1n3c
= lim (lnn)2
nc= 0.
Hencean is convergent, by LCT (b).
Devendra Kumar BITS, Pilani Mathematics I
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Q:46tan 1
n
Devendra Kumar BITS, Pilani Mathematics I
-
Q:46tan 1
n
Sol. Note the series of tan x
tanx= x+ 13x3+ 2
15x5+
Here an = tan 1n . Consider bn = 1n . Note thatbn is
divergent. Now
liman
bn= lim
tan 1n
1n
= 1.
Hencean is divergent, by LCT (a).
Devendra Kumar BITS, Pilani Mathematics I
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Section 10.5
Ratio and Root Tests
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 12. Ratio Test)
Letan be a series of positive terms. Suppose
liman+1an
= r. Then
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 12. Ratio Test)
Letan be a series of positive terms. Suppose
liman+1an
= r. Then(a) If r < 1, the series is convergent;
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 12. Ratio Test)
Letan be a series of positive terms. Suppose
liman+1an
= r. Then(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 12. Ratio Test)
Letan be a series of positive terms. Suppose
liman+1an
= r. Then(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;(c) If r = 1, the test is inconclusive.
Devendra Kumar BITS, Pilani Mathematics I
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Q:32 n lnn
2n.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:32 n lnn
2n.
Sol. Here an = n lnn2n . So
Devendra Kumar BITS, Pilani Mathematics I
-
Q:32 n lnn
2n.
Sol. Here an = n lnn2n . So
an+1an
=(n+1)ln(n+1)
2n+1n lnn2n
= (n+1)ln(n+1)2n lnn
.
Thus
liman+1an
= 12< 1 (on applying LHospitals rule).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:32 n lnn
2n.
Sol. Here an = n lnn2n . So
an+1an
=(n+1)ln(n+1)
2n+1n lnn2n
= (n+1)ln(n+1)2n lnn
.
Thus
liman+1an
= 12< 1 (on applying LHospitals rule).
Therefore by ratio test, the series converges.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:38 n!
nn.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:38 n!
nn.
Sol. Here an = n!nn . So
Devendra Kumar BITS, Pilani Mathematics I
-
Q:38 n!
nn.
Sol. Here an = n!nn . So
an+1an
=(n+1)!
(n+1)n+1n!nn
=( nn+1
)n=
(1 1
n+1
)n
liman+1an
= lim(1 1
n+1
)n= e1 < 1.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:38 n!
nn.
Sol. Here an = n!nn . So
an+1an
=(n+1)!
(n+1)n+1n!nn
=( nn+1
)n=
(1 1
n+1
)n
liman+1an
= lim(1 1
n+1
)n= e1 < 1.
Therefore by ratio test, the series converges.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:42 3n
n32n.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:42 3n
n32n.
Sol. Here an = 3n
n32n. So
Devendra Kumar BITS, Pilani Mathematics I
-
Q:42 3n
n32n.
Sol. Here an = 3n
n32n. So
an+1an
=3n+1
(n+1)32n+13n
n32n
= 32.
n3
(n+1)3
liman+1an
= lim 32.
n3
(n+1)3 =3
2> 1.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:42 3n
n32n.
Sol. Here an = 3n
n32n. So
an+1an
=3n+1
(n+1)32n+13n
n32n
= 32.
n3
(n+1)3
liman+1an
= lim 32.
n3
(n+1)3 =3
2> 1.
Therefore by ratio test, the series diverges.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 13. Root Test or nth Root
Test)
Letan be a series of non-negative terms, and
suppose that lim(an)1n = r. Then
(a) If r < 1, the series is convergent;
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 13. Root Test or nth Root
Test)
Letan be a series of non-negative terms, and
suppose that lim(an)1n = r. Then
(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 13. Root Test or nth Root
Test)
Letan be a series of non-negative terms, and
suppose that lim(an)1n = r. Then
(a) If r < 1, the series is convergent;(b) If r > 1, the series is divergent;(c) If r = 1, the test is inconclusive.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18n2en.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18n2en.
Sol. Here
an = n2en
(an)1/n = n2/ne1
lim(an)1/n = limn2/ne1 = e1 < 1.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:18n2en.
Sol. Here
an = n2en
(an)1/n = n2/ne1
lim(an)1/n = limn2/ne1 = e1 < 1.Therefore the series converges by root test.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:40
n=2n
(lnn)(n/2).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:40
n=2n
(lnn)(n/2).
Sol. Here
an =n
(lnn)(n/2)
(an)1/n = n
1/n
(lnn)1/2
lim(an)1/n = lim n
1/n
(lnn)1/2= 0< 1.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:40
n=2n
(lnn)(n/2).
Sol. Here
an =n
(lnn)(n/2)
(an)1/n = n
1/n
(lnn)1/2
lim(an)1/n = lim n
1/n
(lnn)1/2= 0< 1.
Therefore the series converges by root test.
Devendra Kumar BITS, Pilani Mathematics I
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Section 10.6Alternating Series, Absolute and
Conditional Convergence
Devendra Kumar BITS, Pilani Mathematics I
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Alternating Series
A series whose terms are alternately positive and
negative is called an alternating series.
Devendra Kumar BITS, Pilani Mathematics I
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Alternating Series
A series whose terms are alternately positive and
negative is called an alternating series.
An alternating series is one of the form(1)n+1un
or(1)nun, where un > 0 for all n.
Devendra Kumar BITS, Pilani Mathematics I
-
Examples(1)n+1
Devendra Kumar BITS, Pilani Mathematics I
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Examples(1)n+1(1)n+1 1
n(Alternating harmonic series)
Devendra Kumar BITS, Pilani Mathematics I
-
Examples(1)n+1(1)n+1 1
n(Alternating harmonic series)
(1)n ln(1+ 1
n
)
Devendra Kumar BITS, Pilani Mathematics I
-
Examples(1)n+1(1)n+1 1
n(Alternating harmonic series)
(1)n ln(1+ 1
n
)(1)n n
n+1
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 14. Leibnizs Theorem for
Alternating Series)
The alternating series(1)n+1un converges if
(i) un un+1 for all nN for some N; and
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 14. Leibnizs Theorem for
Alternating Series)
The alternating series(1)n+1un converges if
(i) un un+1 for all nN for some N; and(ii) un 0.
Devendra Kumar BITS, Pilani Mathematics I
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Theorem (Theorem 14. Leibnizs Theorem for
Alternating Series)
The alternating series(1)n+1un converges if
(i) un un+1 for all nN for some N; and(ii) un 0.
Proof. It is enough to show that the sequence Sn is
convergent (definition of convergence of a series). In
order to show that Sn converges, we use the
following result:
Devendra Kumar BITS, Pilani Mathematics I
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Ex. 131 on p. 543: For a sequence {Sn}, if {S2n} and
{S2n+1} converge to the same number L, then Sn L.
Devendra Kumar BITS, Pilani Mathematics I
-
Ex. 131 on p. 543: For a sequence {Sn}, if {S2n} and
{S2n+1} converge to the same number L, then Sn L.First consider the sequence {S2n}. Well see that S2nconverges (by showing that S2n is non-decreasing
and bounded from above).
Devendra Kumar BITS, Pilani Mathematics I
-
Ex. 131 on p. 543: For a sequence {Sn}, if {S2n} and
{S2n+1} converge to the same number L, then Sn L.First consider the sequence {S2n}. Well see that S2nconverges (by showing that S2n is non-decreasing
and bounded from above).
S2n is non-decreasing We have
S2n+2 = S2n+ (u2n+1u2n+2).Since u2n+1u2n+2 0, so S2n+2 S2n. Thus thesequence {S2n} is non-decreasing.
Devendra Kumar BITS, Pilani Mathematics I
-
Ex. 131 on p. 543: For a sequence {Sn}, if {S2n} and
{S2n+1} converge to the same number L, then Sn L.First consider the sequence {S2n}. Well see that S2nconverges (by showing that S2n is non-decreasing
and bounded from above).
S2n is non-decreasing We have
S2n+2 = S2n+ (u2n+1u2n+2).Since u2n+1u2n+2 0, so S2n+2 S2n. Thus thesequence {S2n} is non-decreasing.
Devendra Kumar BITS, Pilani Mathematics I
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S2n is bounded from above Arrange S2n as
S2n = u1 (u2u3) (u2n2u2n1)u2n u1.Thus the sequence {S2n} is bounded from above.
Devendra Kumar BITS, Pilani Mathematics I
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S2n is bounded from above Arrange S2n as
S2n = u1 (u2u3) (u2n2u2n1)u2n u1.Thus the sequence {S2n} is bounded from above.
Therefore it is convergent and so has a limit, say L
i.e., limS2n L.
Devendra Kumar BITS, Pilani Mathematics I
-
Now consider the sequence {S2n+1}. We have
Devendra Kumar BITS, Pilani Mathematics I
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Now consider the sequence {S2n+1}. We have
S2n+1 = S2n+u2n+1 limS2n+1 = limS2n+ limu2n+1.
Devendra Kumar BITS, Pilani Mathematics I
-
Now consider the sequence {S2n+1}. We have
S2n+1 = S2n+u2n+1 limS2n+1 = limS2n+ limu2n+1.
Therefore, using condition (ii):
Devendra Kumar BITS, Pilani Mathematics I
-
Now consider the sequence {S2n+1}. We have
S2n+1 = S2n+u2n+1 limS2n+1 = limS2n+ limu2n+1.
Therefore, using condition (ii):
limS2n+1 = L+0= L.
Devendra Kumar BITS, Pilani Mathematics I
-
Now consider the sequence {S2n+1}. We have
S2n+1 = S2n+u2n+1 limS2n+1 = limS2n+ limu2n+1.
Therefore, using condition (ii):
limS2n+1 = L+0= L.Thus Sn L and hence
(1)n+1un converges to
L.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:8(1)n 10n
(n+1)!.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:8(1)n 10n
(n+1)!.Sol. (i) Here un = 10
n
(n+1)!. Consider
unun+1 =10n
(n+1)!10n+1
(n+2)! =(n8)10n(n+2)! .
Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:8(1)n 10n
(n+1)!.Sol. (i) Here un = 10
n
(n+1)!. Consider
unun+1 =10n
(n+1)!10n+1
(n+2)! =(n8)10n(n+2)! .
Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.(ii) Also, we have limun = lim 10
n
(n+1)! = 0.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:8(1)n 10n
(n+1)!.Sol. (i) Here un = 10
n
(n+1)!. Consider
unun+1 =10n
(n+1)!10n+1
(n+2)! =(n8)10n(n+2)! .
Thus unun+1 0 for all n 8 i.e., un un+1 for alln 8.(ii) Also, we have limun = lim 10
n
(n+1)! = 0.Hence by Laibnizs test the series is convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:12(1)n ln
(1+ 1
n
).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:12(1)n ln
(1+ 1
n
).
Sol. (i) Here un = ln(1+ 1
n
). Consider
f (x)= ln(1+ 1
x
) f (x)= 1
x(x+1).
Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.
Devendra Kumar BITS, Pilani Mathematics I
-
Q:12(1)n ln
(1+ 1
n
).
Sol. (i) Here un = ln(1+ 1
n
). Consider
f (x)= ln(1+ 1
x
) f (x)= 1
x(x+1).
Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.(ii) Also, we have limun = limln
(1+ 1
n
)= 0 (how?).
Devendra Kumar BITS, Pilani Mathematics I
-
Q:12(1)n ln
(1+ 1
n
).
Sol. (i) Here un = ln(1+ 1
n
). Consider
f (x)= ln(1+ 1
x
) f (x)= 1
x(x+1).
Since f (x)< 0 for all x 1. So f (x) is decreasing forall x 1 and hence un un+1, n 1.(ii) Also, we have limun = limln
(1+ 1
n
)= 0 (how?).
Hence by Laibnizs test the series is convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Corollary
For alternating series, if limun does not tend to zero,
then the series diverges.
Devendra Kumar BITS, Pilani Mathematics I
-
Corollary
For alternating series, if limun does not tend to zero,
then the series diverges.
Q:6(1)n+1n2+5
n2+4.
Devendra Kumar BITS, Pilani Mathematics I
-
Corollary
For alternating series, if limun does not tend to zero,
then the series diverges.
Q:6(1)n+1n2+5
n2+4.
Sol. Here un = n2+5
n2+4. Since limun = 1, therefore byabove corollary series is divergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Absolute and Conditional Convergence
A seriesan is said to be absolutely convergent if
the series |an| is convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Absolute and Conditional Convergence
A seriesan is said to be absolutely convergent if
the series |an| is convergent.
If the seriesan is convergent but not absolutely
convergent, then it is said to be conditionally
convergent.
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 16. The Absolute
Convergence Test)
If |an| converges, then an also converges. That is
absolute convergence implies convergence.
Devendra Kumar BITS, Pilani Mathematics I
-
Theorem (Theorem 16. The Absolute
Convergence Test)
If |an| converges, then an also converges. That is
absolute convergence implies convergence.
Proof.We have
|an| an |an| , n.So
0 |an|+an 2 |an| .Now the series
2 |an| converges as the series
|an|converges.
Devendra Kumar BITS, Pilani Mathematics I
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Therefore by Direct Comparison Test, the
non-negative terms series(|an|+an) converges.
Devendra Kumar BITS, Pilani Mathematics I
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Therefore by Direct Comparison Test, the
non-negative terms series(|an|+an) converges.
Now we can writean =
(|an|+an|an|)=
(|an|+an)
|an| .
Thereforean, being a difference of two convergent
series converges.
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
The converse of above theorem is not true.
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
The converse of above theorem is not true.
Example
The alternating harmonic seriesan =
(1)n+1n
converges but |an| = 1n diverges.
Devendra Kumar BITS, Pilani Mathematics I
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Steps to check a series for absolute and conditional
convergence
First check the behavior of the series |an|. If |an| is convergent then an is absolutely
convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Steps to check a series for absolute and conditional
convergence
First check the behavior of the series |an|. If |an| is convergent then an is absolutely
convergent.
If |an| is divergent then check the behavior ofan. If
an is convergent then
an is
conditionally convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Steps to check a series for absolute and conditional
convergence
First check the behavior of the series |an|. If |an| is convergent then an is absolutely
convergent.
If |an| is divergent then check the behavior ofan. If
an is convergent then
an is
conditionally convergent.
Otherwisean is divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:30(1)n lnn
nlnn.
Devendra Kumar BITS, Pilani Mathematics I
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Q:30(1)n lnn
nlnn.Sol.
First we check the behavior of |an| = lnnnlnn . We
havelnn
n lnn lnn
n 1n, n 3.
Also 1
nis divergent so
lnnnlnn is also divergent
(by DCT).
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n lnn
nlnnby using Leibnizs test.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n lnn
nlnnby using Leibnizs test.
(i) Consider f (x)= ln xxlnx f (x)= 1lnx(xlnx)2 . Thus
f (x)< 0 for all x> e and so un un+1 for all n 3.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n lnn
nlnnby using Leibnizs test.
(i) Consider f (x)= ln xxlnx f (x)= 1lnx(xlnx)2 . Thus
f (x)< 0 for all x> e and so un un+1 for all n 3.(ii) Also we have lim lnn
nlnn = 0 (Using LHospitalsrule).
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n lnn
nlnnby using Leibnizs test.
(i) Consider f (x)= ln xxlnx f (x)= 1lnx(xlnx)2 . Thus
f (x)< 0 for all x> e and so un un+1 for all n 3.(ii) Also we have lim lnn
nlnn = 0 (Using LHospitalsrule).
Therefore by Leibnizs test(1)n lnn
nlnn isconvergent.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n lnn
nlnnby using Leibnizs test.
(i) Consider f (x)= ln xxlnx f (x)= 1lnx(xlnx)2 . Thus
f (x)< 0 for all x> e and so un un+1 for all n 3.(ii) Also we have lim lnn
nlnn = 0 (Using LHospitalsrule).
Therefore by Leibnizs test(1)n lnn
nlnn isconvergent.
Hence(1)n lnn
nlnn is conditionally convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:38(1)n+1 (n!)2
(2n)!.
Devendra Kumar BITS, Pilani Mathematics I
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Q:38(1)n+1 (n!)2
(2n)!.
Sol. First we check the behavior of |an| = (n!)2(2n)!.
We have
|an+1||an|
=((n+1)!)2(2n+2)!(n!)2
(2n)!
= (n+1)2
(2n+1)(2n+2),
and so lim|an+1||an| =
14< 1.
Devendra Kumar BITS, Pilani Mathematics I
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Q:38(1)n+1 (n!)2
(2n)!.
Sol. First we check the behavior of |an| = (n!)2(2n)!.
We have
|an+1||an|
=((n+1)!)2(2n+2)!(n!)2
(2n)!
= (n+1)2
(2n+1)(2n+2),
and so lim|an+1||an| =
14< 1.
Therefore by ratio test |an| converges and hence
an is absolutely convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Q:44(1)n 1p
n+pn+1.
Devendra Kumar BITS, Pilani Mathematics I
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Q:44(1)n 1p
n+pn+1.
Sol.
First we check the behavior of |an| = 1p
n+pn+1.
Consider bn = 1pn , then we have
liman
bn= lim
1pn+
pn+1
1pn
= 12.
Therefore by LCT (a) |an| is divergent.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n 1p
n+pn+1 by using Leibnizs test.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n 1p
n+pn+1 by using Leibnizs test.
(i) Since 1pn+
pn+1 >
1pn+1+
pn+2 for all n, so
un > un+1 for all n.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n 1p
n+pn+1 by using Leibnizs test.
(i) Since 1pn+
pn+1 >
1pn+1+
pn+2 for all n, so
un > un+1 for all n.(ii) Also we have limun = lim 1p
n+pn+1 = 0.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n 1p
n+pn+1 by using Leibnizs test.
(i) Since 1pn+
pn+1 >
1pn+1+
pn+2 for all n, so
un > un+1 for all n.(ii) Also we have limun = lim 1p
n+pn+1 = 0.
Therefore by Leibnizs test(1)n 1p
n+pn+1 is
convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Now we check the behavior ofan =
(1)n 1p
n+pn+1 by using Leibnizs test.
(i) Since 1pn+
pn+1 >
1pn+1+
pn+2 for all n, so
un > un+1 for all n.(ii) Also we have limun = lim 1p
n+pn+1 = 0.
Therefore by Leibnizs test(1)n 1p
n+pn+1 is
convergent.
Hence(1)n 1p
n+pn+1 is conditionally convergent.
Devendra Kumar BITS, Pilani Mathematics I
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Summary for Behavior of a Series
If liman9 0, the series diverges.
Devendra Kumar BITS, Pilani Mathematics I
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Summary for Behavior of a Series
If liman9 0, the series diverges.
See if it is known series (like geometric series,
p-series) etc.
Devendra Kumar BITS, Pilani Mathematics I
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Summary for Behavior of a Series
If liman9 0, the series diverges.
See if it is known series (like geometric series,
p-series) etc.
For nonnegative terms series try integral test,
comparison test, ratio test, root test etc.
Devendra Kumar BITS, Pilani Mathematics I
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Summary for Behavior of a Series
If liman9 0, the series diverges.
See if it is known series (like geometric series,
p-series) etc.
For nonnegative terms series try integral test,
comparison test, ratio test, root test etc.
For series with some negative terms (not
necessarily alternating series) see whether |an|
converges. If yes, so doesan.
Devendra Kumar BITS, Pilani Mathematics I
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Summary for Behavior of a Series
If liman9 0, the series diverges.
See if it is known series (like geometric series,
p-series) etc.
For nonnegative terms series try integral test,
comparison test, ratio test, root test etc.
For series with some negative terms (not
necessarily alternating series) see whether |an|
converges. If yes, so doesan.
For alternating series apply Leibnizs test.
Devendra Kumar BITS, Pilani Mathematics I
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Section 10.7
Power Series
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A power series about x= a is a series of the formn=0
an(xa)n in which the center a andcoefficients an are constants.
Devendra Kumar BITS, Pilani Mathematics I
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A power series about x= a is a series of the formn=0
an(xa)n in which the center a andcoefficients an are constants.
A power series about x= 0 is a series which lookslike
n=0
anxn.
Devendra Kumar BITS, Pilani Mathematics I
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A power series about x= a is a series of the formn=0
an(xa)n in which the center a andcoefficients an are constants.
A power series about x= 0 is a series which lookslike
n=0
anxn.
In the expanded form, it looks like
a0+a1x+a2x2+ +anxn+ .Here x is a real variable.
Devendra Kumar BITS, Pilani Mathematics I
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Examplesn=0
xn (geometric series).
Devendra Kumar BITS, Pilani Mathematics I
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Examplesn=0
xn (geometric series).
n=0
xn
n!(exponential series).
Devendra Kumar BITS, Pilani Mathematics I
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Examplesn=0
xn (geometric series).
n=0
xn
n!(exponential series).
(1)n+1 xn
n(logarithmic series).
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Remark.
A power series always converges at the center
i.e.,n=0
an(xa)n converges at x= a andn=0
anxn
converges at x= 0.
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Remark.
A power series always converges at the center
i.e.,n=0
an(xa)n converges at x= a andn=0
anxn
converges at x= 0.So, our aim is to determine the values x, a for
whichn=0
an(xa)n converges.
Devendra Kumar BITS, Pilani Mathematics I
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The following result gives us all the information we
can have regarding convergence/divergence of a
given power series.
Devendra Kumar BITS, Pilani Mathematics I
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The following result gives us all the information we
can have regarding convergence/divergence of a
given power series.
Theorem (Theorem 18. The Convergence
Theorem for Power Series)
If the power seriesn=0
anxn
(i) Converges at x= c(c, 0), then it convergesabsolutely for all x such that |x| < |c|.
Devendra Kumar BITS, Pilani Mathematics I
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The following result gives us all the information we
can have regarding convergence/divergence of a
given power series.
Theorem (Theorem 18. The Convergence
Theorem for Power Series)
If the power seriesn=0
anxn
(i) Converges at x= c(c, 0), then it convergesabsolutely for all x such that |x| < |c|.
(ii) Diverges at x= d, then it diverges for all xsuch that |x| > |d|.
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Remark.
A power seriesan(xa)n can have only one of the
following three features:
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
A power seriesan(xa)n can have only one of the
following three features:
(i) It converges only at x= a and divergeselsewhere.
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
A power seriesan(xa)n can have only one of the
following three features:
(i) It converges only at x= a and divergeselsewhere.
(ii) It converges over an interval i.e., there is a
positive real number R such that series
converges absolutely for |xa| R. The series may ormay not converge at either endpoints
x= aR and x= a+R.
Devendra Kumar BITS, Pilani Mathematics I
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Remark.
A power seriesan(xa)n can have only one of the
following three features:
(i) It converges only at x= a and divergeselsewhere.
(ii) It converges over an interval i.e., there is a
positive real number R such that series
converges absolutely for |xa| R. The series may ormay not converge at either endpoints
x= aR and x= a+R.(iii) It converges absolutely for all x.Devendra Kumar BITS, Pilani Mathematics I
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In case (ii), the interval is called the interval of
convergence of the power series (it may includes one
or both end points) and the half length of the interval
is called the radius of convergence of the power
series (we denote it by R).
Devendra Kumar BITS, Pilani Mathematics I
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Radius of Convergence
A nonnegative number R such that the power seriesn=0
an(xa)n converges absolutely for x such that|xa| R is called theradius of convergence of the power series. The series
may or may not converge at either endpoints
x= aR and x= a+R.
Devendra Kumar BITS, Pilani Mathematics I
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Formula for Radius of Convergence
For the power seriesn=0
an(xa)n the radius ofconvergence is given by the formulae
Devendra Kumar BITS, Pilani Mathematics I
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Formula for Radius of Convergence
For the power seriesn=0
an(xa)n the radius ofconvergence is given by the formulae
1
R= lim
an+1an .
Devendra Kumar BITS, Pilani Mathematics I
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Formula for Radius of Convergence
For the power seriesn=0
an(xa)n the radius ofconvergence is given by the formulae
1
R= lim
an+1an .
or1
R= lim |an|1/n .
Devendra Kumar BITS, Pilani Mathematics I
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Formula for Radius of Convergence
For the power seriesn=0
an(xa)n the radius ofconvergence is given by the formulae
1
R= lim
an+1an .
or1
R= lim |an|1/n .
Remark.
If series converges only at center then R = 0.
Devendra Kumar BITS, Pilani Mathematics I
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Formula for Radius of Convergence
For the power seriesn=0
an(xa)n the radius ofconvergence is given by the formulae
1
R= lim
an+1an .
or1
R= lim |an|1/n .
Remark.
If series converges only at center then R = 0.If series converges for all x then R =.
Devendra Kumar BITS, Pilani Mathematics I