Lecture 3 & 4 - uniba.sk · 2019-12-28 · 11.10.2015 1 Lecture 3 & 4 Infinite sequences and series...
Transcript of Lecture 3 & 4 - uniba.sk · 2019-12-28 · 11.10.2015 1 Lecture 3 & 4 Infinite sequences and series...
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Lecture 3 & 4
Infinite sequences and series
Sequences
Infinite sequences are functions whose domain is the set of natural numbers
ℕ = {1, 2, 3, 4, . . . , 𝑛, . . . }.
Definition. Infinite sequence is a function whose domain is the set of natural
numbers ℕ, which assigns to each number 𝑛 ∈ ℕ exactly one real number 𝑎𝑛 ∈ ℝ,
such that [𝑛, 𝑎𝑛 ] ∈ 𝑝:
𝑛 ∈ ℕ ∃! (𝑎𝑛 ∈ ℝ ) ([𝑛, 𝑎𝑛 ] ∈ 𝑝)
Notation of sequences:
{𝑎𝑛 }𝑛=1∞ = 𝑎𝑛 = {𝑎1, 𝑎2, 𝑎3, … }
𝑎𝑛 is called nth member of the sequence, 𝑛 - corresponding independent variable
Graph of a sequence: isolated points, Fig. 2.1.
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Fig. 2.1. Graph of sequence 𝑎𝑛 = 1
𝑛 . Contains points [𝑛, 𝑎𝑛 ], 𝑛 ∈ ℕ, 𝑎𝑛 ∈ ℝ.
Sequence can be defined:
listing of several initial members
defining the rule for calculation of 𝑎𝑛 ,
recurrently - 𝑎𝑛+1 is defined from 𝑎𝑛
e.g.
𝑛2 = 1, 4, 9, 16, …
𝑛
𝑛+1 =
1
2,
2
3,
3
4,
4
5, …
2𝑛+(−1)𝑛
𝑛 = 1,
5
2,
5
3,
9
4,
9
5,
13
6, …
𝑛
𝑎𝑛
2
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Example. Put down first 5 members of a sequence whose nth member is given as:
𝑎𝑛 =𝑛+2
2𝑛+1.
𝑛 = 1, … ,5 𝑎𝑛 = 1,4
5,
5
7,
6
9,
7
11, …
Example. Find the formula for nth member of sequence:
𝑏𝑛 = 0, −3
2, −
8
3, −
15
4, −
24
5, … .
Written in the form of compound fractions:
𝒏 1 2 3 4 5
𝒃𝒏 0 −11
2 −2
2
3 −3
3
4 −4
4
5
𝑎𝑛 decreases with increasing n, but is by 1
𝑛 larger than −𝑛: 𝑎𝑛 = −𝑛 +
1
𝑛=
1
𝑛− 𝑛. The
formula holds also for 𝑛 = 1.
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Example. Put down first 6 members of a sequence given by recursive formula:
𝑎1 = 𝑎2 = 1, 𝑎𝑛+1 = 𝑎𝑛−1 + 𝑎𝑛
Solve by substituting numbers:
𝑎1 = 1
𝑎2 = 1
𝑎3 = 𝑎1 + 𝑎2 = 1 + 1 = 2
𝑎4 = 𝑎2 + 𝑎3 = 1 + 2 = 3
𝑎5 = 𝑎3 + 𝑎4 = 2 + 3 = 5
𝑎6 = 𝑎4 + 𝑎5 = 3 + 5 = 8
Then: 𝑎𝑛 = {1, 1, 2, 3, 5, 8, … }
It is called sequence of Fibonacci1
________________________
1 Leonardo Pisano (1170-1250), known as Fibonacci, was Italian mathematician
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Arithmetic sequence:
𝑎𝑛+1 = 𝑎𝑛 + 𝑑
Number d is difference. Calculation of 𝑎𝑛 from 𝑑:
𝑎𝑛 = 𝑎1 + 𝑛 − 1 𝑑 𝑎𝑛 = 𝑎𝑚 + (𝑛 − 𝑚)𝑑
Sum 𝑆𝑛 of first 𝑛 members of arithmetic sequence:
𝑆𝑛 =𝑛
2(𝑎1 + 𝑎𝑛)
Example. Calculate the sum of first 15 members of arithmetic sequence, if:
𝑎4 = 9 a 𝑎7 = 15
Calculate 𝑑 from 𝑎4 and 𝑎7:
𝑎4: 𝑎1 + 3𝑑 = 9 𝑎7: 𝑎1 + 6𝑑 = 15
We obtain: 𝑎1 = 3 and 𝑑 = 2. Then 𝑎15 = 𝑎1 + 14𝑑 = 31
Sum:
𝑆15 =𝑛
2 𝑎1 + 𝑎𝑛 =
15
2 3 + 31 = 255
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Geometric sequence is defined by recurrent formula:
𝑎𝑛+1 = 𝑎𝑛𝑞
number q - quotient.
Calculation of 𝑎𝑛 :
𝑎𝑛 = 𝑎1𝑞𝑛−1 𝑎𝑛 = 𝑎𝑚𝑞𝑛−𝑚
Sum 𝑆𝑛 of first 𝑛 members of geometric sequence with q 1 and 𝑎 ≡ 𝑎1:
𝑆𝑛 = 𝑎 + 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + … + 𝑎𝑞𝑛−1
𝑞𝑆𝑛 = 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + 𝑎𝑞4 + … + 𝑎𝑞𝑛
𝑆𝑛 − 𝑞𝑆𝑛 = 𝑎 − 𝑎𝑞𝑛
𝑆𝑛 =𝑎(1−𝑞𝑛 )
1−𝑞
Graphs of geometric sequences depending on quotient q, Fig. 2.2.
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A B
C D
Fig. 2.2. Graph of geometric sequence 𝑎𝑛 = 𝑎1𝑞𝑛−1 for 𝑎1 = 1 and various
values of 𝑞. A. 𝑞 = −1,2 B. 𝑞 = −0,8 C. 𝑞 = 0,8 D. 𝑞 = 1,2.
𝑞 = 1,2
𝑛 𝑛
𝑎𝑛 𝑎𝑛 𝑞 = 0,8
𝑛 𝑛
𝑎𝑛
𝑎𝑛 𝑞 = −1,2 𝑞 = −0,8
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Example. Calculate the sum of first 10 members of geometric sequence, if:
𝑏5 =2
81 and 𝑏7 =
2
729
First we calculate constants 𝑏1 a 𝑞 from 𝑏5 a 𝑏7 :
𝑏5: 𝑏1𝑞4 =2
81 𝑏7: 𝑏1𝑞6 =
2
729
Solution: 𝑏1 = 2 and 𝑞 =1
3, then the sum 𝑆10:
𝑆10 = 𝑏11−𝑞10
1−𝑞= 2
1−1
310
1−1
3
= 2,99995
With increasing value of independent variable n the values of corresponding
members 𝑎𝑛 can increase, decrease or converge to certain number.
Sequence is called increasing [decreasing, non-decreasing or non-increasing] if for
each 𝑛 ∈ ℕ holds:
𝑎𝑛 < 𝑎𝑛+1 [𝑎𝑛 > 𝑎𝑛+1, 𝑎𝑛 ≤ 𝑎𝑛+1, 𝑎𝑛 ≥ 𝑎𝑛+1].
Sequence {𝑎𝑛}𝑛=1∞ is called bounded below [above] if for each 𝑛 ∈ ℕ exists 𝐻 ∈ ℝ,
such that: 𝑎𝑛 > 𝐻 (𝑎𝑛 < 𝐻).
Sequence is called bounded if it is bounded below and above.
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Example. Prove that sequence 𝑛+1
𝑛+2 𝑛=1
∞ is growing and bounded above.
For increasing sequence: 𝑎𝑛 < 𝑎𝑛+1
𝑛+1
𝑛+2<
𝑛+1 +1
𝑛+1 +2
𝑛+1
𝑛+2<
𝑛+2
𝑛+3
𝑛 + 1 𝑛 + 3 < (𝑛 + 2)2
𝑛2 + 4𝑛 + 3 < 𝑛2 + 4𝑛 + 4
3 < 4
therefore: sequence is increasing
For 𝑛 = 1 we get the lower bound 𝐻𝑙 = 𝑎1 =2
3 , therefore: 𝑎𝑛 ≥ 𝐻𝑠 =
2
3.
We estimate the upper bound as: 𝐻𝑢 = 1.
𝑛+1
𝑛+2< 1
𝑛 + 1 < 𝑛 + 2
1 < 2
Therefore 𝑛+1
𝑛+2 𝑛=1
∞ is increasing and bounded: 𝐻𝑙 =
2
3 and 𝐻𝑢 = 1.
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Limit of sequence
Graphs of sequences on Fig. 2.3. show sequence {𝑛2} which with increasing value
of 𝑛 increases beyond any bounds (sequence is divergent). On the other hand,
sequence 1
𝑛2 is with increasing 𝑛 approaching zero (sequence is converging
to zero).
A B
Fig. 2.3. Graphs of infinite sequences. A. 𝑎𝑛 = 𝑛2 . B. 𝑏𝑛 = 1
𝑛2 .
𝑛 𝑛
𝑎𝑛 𝑏𝑛
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Closeness to a point is related to unlimited approaching.
𝜀-environment of a point 𝑏, 𝑂𝜀(𝑏), is defined as the interval (𝑏 − 𝜀, 𝑏 + 𝜀) with a
center in 𝑏 and width 𝜀 > 0, or as a set 𝑂𝜀 𝑏 = {𝑥 ∈ ℝ; |𝑏 − 𝑥| < 𝜀}. An infinite
sequence is approaching its limit 𝑏 if for any small 𝜀 points 𝑎𝑛 lie within the
environment 𝑂𝜀(𝑏), Fig. 2.4:
∀𝜀 ∈ ℝ+, 𝜀 << 1 ∃𝑛𝜀 ; 𝑛 > 𝑛𝜀 ⇒ |𝑎𝑛– 𝑏| < 𝜀
Fig. 2.4. Graph of sequence 𝑎𝑛 = 3𝑛+2
𝑛+1 . Pre 𝑛 ≥ 5 its points 𝑎𝑛 lie within a
narrow band around number 3 (3 − 𝜀, 3 + 𝜀), 0 < 𝜀 ≪ 1.
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1 Symbol „≪“ in the inequality 𝑐 ≪ 𝑑 means 𝑐 is much smaller than 𝑑.
𝑛
𝑎𝑛
𝑂𝜀(𝑏)
𝑛𝜀
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Definition. We say that number 𝑏 is the limit of sequence {𝑎𝑛}𝑛=1∞ , if for all é 𝜀 >
0, 𝜀 ∈ ℝ exists 𝑛𝜀 ∈ ℕ, that for 𝑛 > 𝑛𝜀 :
𝑏 − 𝜀 < 𝑎𝑛 < 𝑏 + 𝜀 or |𝑏 − 𝑎𝑛 | < 𝜀
we write: lim𝑛→∞ 𝑎𝑛 = 𝑏
and say „limit of 𝑎𝑛 for 𝑛 increasing to infinity converges to the finite number 𝑏“.
Such sequence is called convergent. Sequence which does not have a limit is
divergent.
Example. Prove that the limit of sequence 3𝑛+2
𝑛+1 from Fig. 1.4., is equal to:
lim𝑛→∞3𝑛+2
𝑛+1= 3.
By definition for all 𝑛 > 𝑛𝜀 and 𝜀 > 0:
3 − 𝜀 <3𝑛+2
𝑛+1 < 3 + 𝜀
Right hand side: 3𝑛+2
𝑛+1 < 3 + 𝜀 after rearrangement: 0 < 𝜀𝑛 + 𝜀 + 1, which is true
for all 𝑛 ∈ ℕ.
Left hand side: 3 − 𝜀 <3𝑛+2
𝑛+1 can be converted to:
1
𝑛+1< 𝜀 shows for what 𝑛𝜀 the
inequality hold if we choose parameter 𝜀:
𝜀 𝟎, 𝟑𝟑 𝟎, 𝟏𝟓 𝟎, 𝟎𝟗 𝟎, 𝟎𝟓 𝟎, 𝟎𝟏
𝑛𝜀 2 5 10 20 100
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Example. Prove that the limit of sequence 1
𝑛 is equal to: lim𝑛→∞
1
𝑛= 0.
If the limit is equal to 0, then:
0 − 𝜀 <1
𝑛 < 0 + 𝜀
- left hand side is always true: 𝑛 > −1
𝜀 for 𝑛 ∈ ℕ.
- right hand side: 𝑛 >1
𝜀 for any 𝜀 we can always find: 𝑛 > 𝑛𝜀 >
1
𝜀 so that the this
inequality is fulfilled. Therefore: lim𝑛→∞1
𝑛= 0.
Theorem. Limit of sequence 1
𝑛𝑘 is equal to: lim𝑛→∞1
𝑛𝑘= 0 where 𝑘 ∈ ℕ
Existence of a limit:
Theorem. Each bounded monotonic sequence1 has a limit. Each sequence has at
most one limit.
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1 The term monotonic function or sequence means that a curve in its domain only increases or
only decreases and does not combine increase with a decrease.
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Definition. The sequence {𝑎𝑛}𝑛=1∞ has a infinite limit of +∞ (−∞), if for each
𝐺 ∈ ℝ exists 𝑛 such, that for all 𝑛 > 𝑛 holds: 𝑎𝑛 > 𝐺 (𝑎𝑛 < 𝐺), which can be
written as:
lim𝑛→∞ 𝑎𝑛 = ∞ (lim𝑛→∞ 𝑎𝑛 = −∞)
Properties of limits:
Theorem. Let us have two convergent sequences 𝑎𝑛 and 𝑏𝑛 , which have limits
equal to lim𝑛→∞ 𝑎𝑛 = 𝑎, lim𝑛→∞ 𝑏𝑛 = 𝑏
lim𝑛→∞(𝑎𝑛 + 𝑏𝑛 ) = 𝑎 + 𝑏
lim𝑛→∞(𝑎𝑛 − 𝑏𝑛 ) = 𝑎 − 𝑏
lim𝑛→∞ 𝑘 ⋅ 𝑎𝑛 = 𝑘 ⋅ 𝑎 kde 𝑘 ∈ ℝ
lim𝑛→∞ |𝑎𝑛 | = |𝑎|
lim𝑛→∞( 𝑎𝑛 ⋅ 𝑏𝑛 ) = 𝑎 ⋅ 𝑏
lim𝑛→∞𝑎𝑛
𝑏𝑛=
𝑎
𝑏, ak 𝑏𝑛 , 𝑏 ≠ 0
lim𝑛→∞ 𝑏𝑎𝑛 = 𝑏𝑎 ak 𝑏 ∈ ℝ, 𝑏 > 0
lim𝑛→∞ 𝑎𝑛𝑏 = 𝑎𝑏 ak 𝑏 ∈ ℝ, 𝑎 > 0
lim𝑛→∞ 1 +1
𝑛
𝑛= 𝑒
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Example. Determine the limit of sequence: lim𝑛→∞2𝑛3+3𝑛2−𝑛+1
5𝑛3−2𝑛2+4𝑛−7.
We convert this limit of the rational function to a collection of limits of the type
lim𝑛→∞1
𝑛𝑘, by multiplying it with a factor:
1/𝑛𝑚
1/𝑛𝑚, where 𝑚 is the highest power 𝑛𝑚
occurring in the fraction.
lim𝑛→∞2𝑛3+3𝑛2−𝑛+1
5𝑛3−2𝑛2+4𝑛−7= lim𝑛→∞
2𝑛3+3𝑛2−𝑛+1
5𝑛3−2𝑛2+4𝑛−7⋅
1
𝑛3
1
𝑛3
= lim𝑛→∞
2+3
𝑛−
1
𝑛2+1
𝑛3
5−2
𝑛+
4
𝑛2−7
𝑛3
=
=lim 𝑛→∞ 2+lim 𝑛→∞
3
𝑛−lim 𝑛→∞
1
𝑛2+lim 𝑛→∞1
𝑛3
lim 𝑛→∞ 5−lim 𝑛→∞2
𝑛+lim 𝑛→∞
4
𝑛2−lim 𝑛→∞7
𝑛3
=2+0−0+0
5−0+0−0=
2
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Example. Determine the limit of sequence: lim𝑛→∞ 3𝑛+2
2𝑛−1
2+1
𝑛.
lim𝑛→∞ 3𝑛+2
2𝑛−1
2+1
𝑛= lim𝑛→∞
3𝑛+2
2𝑛−1
lim 𝑛→∞ 2+1
𝑛
= 3
2
2=
9
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Example. Determine the limit of sequence: lim𝑛→∞ 𝑛+1
𝑛
𝑛2
3𝑛3+1.
lim𝑛→∞ 𝑛+1
𝑛
𝑛2
3𝑛3+1= lim𝑛→∞ 1 +
1
𝑛
𝑛⋅𝑛
3𝑛3+1= 𝑒
lim 𝑛→∞𝑛
3𝑛3+1 = 𝑒0
3+0 = 𝑒0 = 1
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The theorem characterizing the properties of limits can be cautiously applied also to
infinite limits. For:
lim𝑛→∞ 𝑎𝑛 = ∞ and lim𝑛→∞ 𝑏𝑛 = ∞
then: lim𝑛→∞(𝑎𝑛 + 𝑏𝑛) = ∞ which can be written as: ∞ + ∞ = ∞.
In a simplified notation:
∞ + ∞ = ∞, −∞ −∞ = −∞
∞ ± 𝑐 = ∞, −∞ ± 𝑐 = −∞ where 𝑐 ∈ ℝ
𝑐 ⋅∞ = ∞, 𝑐 ⋅ −∞ = −∞ where 𝑐 ∈ ℝ, 𝑐 > 0
𝑑 ⋅∞ = −∞, 𝑑 ⋅ −∞ = ∞ where 𝑑 ∈ ℝ, 𝑑 < 0
𝑐∞ = ∞, 𝑐−∞ = 0 where 𝑐 ∈ ℝ, 𝑐 > 1
𝑑∞ = 0, 𝑑−∞ = ∞ where 𝑑 ∈ ℝ, 0 < 𝑑 < 1
𝑐
∞= 0,
𝑐
−∞= 0 where 𝑐 ∈ ℝ
When evaluating limits we can find undefined expressions:
0 ⋅∞, 0 ⋅ −∞ , ∞ −∞,0
0,
±∞
±∞, 1∞, 1−∞ these must be solved individually.
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Infinite series
Let 𝑎𝑛 𝑛=1∞ = 1,
1
3,
1
9,
1
27, … ,
1
3𝑛−1, … be an infinite sequence with 𝑎𝑛 =
1
3𝑛−1
(𝑛 ∈ ℕ). With help of 𝑎𝑛 we will form a new sequence {𝑆𝑛} with members equal
to the sum of first 𝑛 members of 𝑎𝑛 :
𝑆1 = 1
𝑆2 = 1 +1
3= 1
1
3
𝑆3 = 1 +1
3+
1
9= 1
4
9
𝑆4 = 1 +1
3+
1
9+
1
27= 1
13
27
𝑆𝑛 = 1 +1
3+
1
9+
1
27+ ⋯ +
1
3𝑛−1=
1
3𝑛−1𝑛1
It can be shown that limit of this sequence {𝑆𝑛} obtained as partial sums of the
sequence 𝑎𝑛 will have for 𝑛 → ∞ the shape:
𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎𝑛 + … = 𝑎𝑛∞𝑛=1
The limit in the example shown above of the sequence 𝑎𝑛 =1
3𝑛−1 will be equal to:
lim𝑛→∞ 𝑆𝑛 =3
2.
However sum of an infinite series may not exist (can be infinite). It exists only if the
sequence of partial sums converges i.e. has a finite limit.
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Theorem. We say that infinite series 𝑎𝑛∞𝑛=1 is convergent (has a finite sum), if
the sequence of partial sums deriving from it {𝑆𝑛} is convergent and has a limit:
lim𝑛→∞ 𝑆𝑛 = 𝑆. Then also the sum of the infinite series equal to 𝑆:
𝑎𝑛∞𝑛=1 = lim𝑛→∞ 𝑆𝑛 = 𝑆
Example . Calculate the sum of infinite series: 1 +1
2+
1
4+
1
8+
1
16+ … =
1
2𝑘∞𝑘=0 .
First we form the series of partial sums:
𝑆𝑘 = 1
2𝑘0𝑘=0 ,
1
2𝑘1𝑘=0 ,
1
2𝑘2𝑘=0 ,
1
2𝑘3𝑘=0 , … =
= 1, 1 +1
2, 1 +
1
2+
1
4, 1 +
1
2+
1
4+
1
8, … = 1,
3
2,
7
4,
15
8, … = 2 −
1
2𝑘−1
The sum exist if 𝑆𝑘 = {2 −1
2𝑘−1} converges:
lim𝑘→∞ 2 −1
2𝑘−1 = 2 − lim𝑘→∞1
2𝑘−1= 2 − 0 = 2
Then:
𝑆 = 1
2𝑘∞𝑘=0 = lim
𝑘→∞ 𝑆𝑘 = lim𝑘→∞ 2 −
1
2𝑘−1 = 2
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Theorem. Let 𝑎𝑛∞𝑛=1 be an infinite series and suppose the limit: lim𝑛→∞
𝑎𝑛+1
𝑎𝑛 =
𝐿 exists (D’Alembert criterion of convergence)1
If 𝐿 < 1 then the series converges, if 𝐿 > 1 then the series diverges, if 𝐿 = 1
then it is not possible to decide based on this criterion.
Example. Decide whether the infinite series: 𝑛
𝑒𝑛∞𝑛=1 converges or not.
D’Alembert criterion:
𝐿 = lim𝑛→∞ 𝑎𝑛+1
𝑎𝑛 = lim𝑛→∞
𝑛+1
𝑒𝑛+1𝑛
𝑒𝑛 = lim𝑛→∞
𝑛+1 𝑒𝑛
𝑛𝑒𝑛+1 =
1
e⋅ lim𝑛→∞
𝑛+1
𝑛=
=1
e⋅ lim𝑛→∞
1+1
𝑛
1= =
1
e⋅
1+0
1=
1
e< 1
Since 𝐿 < 1 the series converges. The sum is, however, unknown.
_________________________
1 Jean le Rond D'Alembert (1717-1783) French mathematician
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Example. Decide whether the infinite series: 1
𝑛 !∞𝑛=1 converges or not.
D’Alembert criterion:
𝐿 = lim𝑛→∞ 𝑎𝑛+1
𝑎𝑛 = lim𝑛→∞
1
𝑛+1 !1
𝑛 !
= lim𝑛→∞ 𝑛 !
𝑛+1 ! = lim𝑛→∞
𝑛 !
𝑛+1 ⋅𝑛 ! =
= lim𝑛→∞1
𝑛+1= lim𝑛→∞
1
𝑛
1+1
𝑛
=0
1+0= 0 < 1
Since 𝐿 < 1 the series converges.
The sum of an infinite series can be obtained for geometric infinite series.
The sum of first n members of a geometric sequence {𝑎𝑞𝑛} is: 𝑆𝑛 =𝑎(1−𝑞𝑛 )
1−𝑞.
Infinite geometric series is the sum of infinite number of the members of geometric
sequence:
𝑎 + 𝑎𝑞 + 𝑎𝑞2 + 𝑎𝑞3 + … + 𝑎𝑞𝑛 + … = 𝑎𝑞𝑛∞𝑛=0
its sum is defined as the limit 𝑆𝑛 :
𝑆 = 𝑎𝑞𝑛∞𝑛=0 = lim𝑛→∞ 𝑆𝑛 = lim𝑛→∞
𝑎(1−𝑞𝑛 )
1−𝑞=
𝑎
1−𝑞⋅ lim𝑛→∞ 1 − 𝑞𝑛
If the quotient q:
|𝑞| < 1 then lim𝑛→∞ 1 − 𝑞𝑛 = 1 exists: 𝑆 = 𝑎𝑞𝑛∞𝑛=0 =
𝑎
1−𝑞
|𝑞| > 1 then lim𝑛→∞ 1 − 𝑞𝑛 does not exist and the sum of the infinite
geometric series diverges (is infinite)
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21
Example. Determine whether infinite series: 2 −1
4
𝑛∞𝑛=0 converges and if yes,
then find its sum:
Geometric series, coefficient: 𝑎 = 2 and quotient: 𝑞 = −1
4, −
1
4 =
1
4< 1,
Therefore series converges and its sum is:
2(−1
4)𝑛∞
𝑛=0 =𝑎
1−𝑞=
2
1−(−1
4)
=8
5
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Power series
So far we have considered infinite series whose members were constants (numbers)
𝑎𝑛 ∈ ℝ:
𝑎1 + 𝑎2 + 𝑎3 + … + 𝑎𝑛 + … = 𝑎𝑛∞𝑛=1
Power series are composed of terms which contain coefficients an as well as the
independent variable 𝑥 and its powers 𝑥𝑛 :
𝑎0 + 𝑎1𝑥 + 𝑎2𝑥2 + … + 𝑎𝑛𝑥𝑛 + … = 𝑎𝑛∞𝑛=0 𝑥𝑛
where: 𝑛 ∈ ℕ
Power series - polynomial of infinite degree (𝑛 → ∞)
By inserting a value of independent variable: 𝑥 = 𝑟 power series turns to infinite
numerical series: 𝑎𝑛∞𝑛=0 𝑟𝑛 (which may converge for the given 𝑟).
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Example. Find the sum of power series: 𝑥𝑛∞𝑛=0 .
For this series: 𝑎0 = 𝑎1 = ⋯ = 𝑎𝑛 = … = 1
It represents a after insertion 𝑥 = 𝑟 a geometric series with 𝑎 = 1 and quotient r.
If |𝑟| < 1 then this series converges to the sum 𝑆:
𝑆 = 𝑟𝑛∞𝑛=0 =
1
1−𝑟.
if for example 𝑟 =1
2 then 𝑆 =
1
2
𝑛∞𝑛=0 =
1
1−1
2
= 2.
if for example 𝑟 = −2
3 then 𝑆 = −
2
3
𝑛∞𝑛=0 =
1
1+2
3
=3
5
This power series represents a function, which assigns to each 𝑟 (|𝑟| < 1) a number: 1
1−𝑟. Thus, we say, the power series: 𝑥𝑛∞
𝑛=0 converges to a function: 𝑓 𝑥 =1
1−𝑥
on the interval: 𝑥 ∈ (−1, 1).
Each power series converges on a symmetric interval with the center in zero:
(−𝑟, 𝑟) a the parameter 𝑟 is called: radius of convergence.
Radius of convergence of a power series: 𝑎𝑛∞𝑛=0 𝑥𝑛 can be determined as:
𝑟 = lim𝑛→∞ 𝑎𝑛
𝑎𝑛+1
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Example. Find a radius of convergence of power series:
𝑥
2
𝑛∞𝑛=0 = 1 +
1
2𝑥 +
1
4𝑥2 +
1
8𝑥3 + … +
1
2𝑛𝑥𝑛 + ….
Radius: 𝑟 = lim𝑛→∞ 𝑎𝑛
𝑎𝑛+1 = lim𝑛→∞
1
2𝑛
1
2𝑛+1
= lim𝑛→∞ 2 = 2
The power series 𝑥
2
𝑛∞𝑛=0 converges on interval: (−2, 2).
Since it is a geometric series with quotient 𝑞 =𝑥
2, we can determine its sum and find
a function to which it converges on (−2, 2):
𝑆 = 𝑥
2
𝑛∞𝑛=0 =
1
1−𝑞=
1
1−𝑥
2
=2
2−𝑥= 𝑓(𝑥)
Power series exists, which have the interval of convergence of (−∞,∞), e.g.:
2𝑛
𝑛 !𝑥𝑛∞
𝑛=0 .
If we know the sum of a power series we can replace it by a function 𝑆(𝑥).
Conversely, in some situations it may be convenient to expand a function 𝑆 𝑥 into
the form of a power series:
𝑆 𝑥 = 𝑎𝑛𝑥𝑛∞𝑛=0 .