Section 9.5b

Testing Convergenceat Endpoints

Alternating SeriesA series in which the terms are alternatingly positive andnegative is an alternating series.

Examples: 1 41 1 12 1

2 4 8 2

n

n

This is a geometric series with a = –2 and r = –1/2

So it converges to: 2

1 1 2

23 2

43

Alternating SeriesA series in which the terms are alternatingly positive andnegative is an alternating series.

Examples: 11 2 3 4 5 6 1 n n This series diverges by the nth-Term Test:

1lim 1 n

nn

111 1 1 112 3 4 5

n

n

This is the alternating harmonic series, and it converges bya new test that we will see shortly…

The Alternating Series Test (Leibniz’s Thm)

The series 11 2 3 4

1

1 nn

n

u u u u u

converges if all three of the following conditions are satisfied:1. each is positive;nu2. for all , for some integer ;1n nu u n N N3. .lim 0nn

u

The Alternating Series Estimation ThmIf the alternating series satisfies the conditions of Leibniz’s Theorem, then the truncation errorfor the nth partial sum is less than and has the samesign as the first unused term.

1

11 n

nnu

1nu

The Alternating Series Test (Leibniz’s Thm)Closing in on the sum of a convergent alternating series:

x0

2S 4S 1S3S

1u

L

2u3u

4u

This figure not only proves the fact of convergence; it alsoshows the way that an alternating series converges. Thepartial sums keep “overshooting” the limit as they go backand forth on the number line, gradually closing in as theterms tend to zero.

The Alternating Series Test (Leibniz’s Thm)Prove that the alternating harmonic series is convergent,but not absolutely convergent. Find a bound for thetruncation error after 99 terms.

1 11 .2 3

The terms are strictly alternating in sign and decrease inabsolute value from the start:

1lim 0n n

Also,

By the Alternating Series Test, 1

1

1 n

n n

converges.

The Alternating Series Test (Leibniz’s Thm)Prove that the alternating harmonic series is convergent,but not absolutely convergent. Find a bound for thetruncation error after 99 terms.

However, the series

of absolute values is the harmonic series, which we knowdiverges, so the alternating harmonic series is notabsolutely convergent.

1

1n n

The Alternating Series Estimation Theorem guaranteesthat the truncation error after 99 terms is less than

99 11 1

99 1 100u

Absolute and Conditional ConvergenceBecause the alternating harmonic series is convergent butnot absolutely convergent, we say it is conditionallyconvergent (or converges conditionally).

We take it for granted that we can rearrange the term of a finite sum without affecting the sum. We can also rearrange a finite number of terms of an infinite series without affecting the sum. But if we rearrange an infinite number of terms of an infinite series, we can be sure of leaving the sum unaltered only if it converges absolutely.

Practice ProblemsDetermine whether the given series converges absolutely,converges conditionally, or diverges. Give reasons for youranswer. 1

21

11 n

n

nn

is a decreasing series with positive terms, and2

1n

nun

lim 0nnu

So the series converges by the Alternating Series Test.

Check for absolute convergence: 21

1n

nn

Practice ProblemsDetermine whether the given series converges absolutely,converges conditionally, or diverges. Give reasons for youranswer. 1

21

11 n

n

nn

behaves like for large values of n…21

1n

nn

1n

1

1n n

diverges, so also diverges…2

1

1n

nn

Conclusion:

The original series converges conditionally

Practice ProblemsDetermine whether the given series converges absolutely,converges conditionally, or diverges. Give reasons for youranswer. 1

21

sin1 n

n

nn

Take a look at:2

1

sinn

nn

Conclusion:

The original series converges absolutely

is always less thanor equal to: 2

1

1n n

Which we know convergesas a p-series with p = 2.

So this series also convergesby direct comparison

Intervals of ConvergenceHow to Test a Power Series for Convergence

0

nn

n

c x a

1. Use the Ratio Test to find the values of x for which theseries converges absolutely. Ordinarily, this is an openinterval .a R x a R In some instances, the series converges for all values of x.In rare cases, the series converges only at x = a.

2. If the interval of convergence is finite, test forconvergence or divergence at each endpoint. The RatioTest fails at these points. Use a comparison test, the Integral Test, or the Alternating Series Test.

Intervals of ConvergenceHow to Test a Power Series for Convergence

0

nn

n

c x a

3. If the interval of convergence is ,conclude that the series diverges (it does not evenconverge conditionally) for , because for thosevalues of x the nth term does not approach zero.

a R x a R

x a R

Finally, take a look at the flowchart on p.505,and read the “Word of Caution”…

Practice ProblemsFind (a) the interval of convergence of the series. For what values ofx does the series converge (b) absolutely, (c) conditionally?

0

5 n

n

x

This is a geometric series which converges only for

5 1x

1 5 1x

6 4x (a) 6, 4

(b) 6, 4 (c) None

Practice ProblemsFind (a) the interval of convergence of the series. For what values ofx does the series converge (b) absolutely, (c) conditionally?

1

3 2 n

n

xn

Ratio Test for absolute convergence:

1lim n

nn

aa

13 2lim

1 3 2

n

nn

x nn x

3 2x

The series converges absolutely when 3 2 1x 1 3 2 1x 1 13

x Now, we check the endpoints…

Practice ProblemsFind (a) the interval of convergence of the series. For what values ofx does the series converge (b) absolutely, (c) conditionally?

1

3 2 n

n

xn

Check

1 :3

x

1

1 n

n n

converges conditionally

Check 1:x 1

1n

n n

diverges

(a)1 ,13

(b)1 ,13

(c) At13

x

Practice ProblemsFind (a) the interval of convergence of the series. For what values ofx does the series converge (b) absolutely, (c) conditionally?

2 1

0 !

n

n

xn

Ratio Test for absolute convergence:

1lim n

nn

aa

2 3

2 1

!lim1 !

n

nn

x nn x

2

lim1n

xn

0

This series converges absolutely for all real numbers.

(a) All real numbers(b) All real numbers(c) None