Lecture 27 :Alternating Series - University of Notre Dameapilking/Math10560/Lectures/Lecture...

Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation

Alternating Series

The integral test and the comparison test given in previous lectures, apply onlyto series with positive terms.

I A series of the formP∞

n=1(−1)nbn orP∞

n=1(−1)n+1bn, where bn > 0 forall n, is called an alternating series, because the terms alternate betweenpositive and negative values.

I Example∞Xn=1

(−1)n

n= −1 +

2− 1

4− 1

5+ . . .

∞Xn=1

(−1)n+1 n

2n + 1=

3− 2

7− 4

9+ . . .

I We can use the divergence test to show that the second series abovediverges, since

limn→∞

(−1)n+1 n

2n + 1does not exist

Annette Pilkington Lecture 27 :Alternating Series

Alternating Series

n=1(−1)nbn orP∞

I Example∞Xn=1

(−1)n

n= −1 +

2− 1

4− 1

5+ . . .

∞Xn=1

(−1)n+1 n

2n + 1=

3− 2

7− 4

9+ . . .

limn→∞

(−1)n+1 n

Alternating Series

n=1(−1)nbn orP∞

I Example∞Xn=1

(−1)n

n= −1 +

2− 1

4− 1

5+ . . .

∞Xn=1

(−1)n+1 n

2n + 1=

3− 2

7− 4

9+ . . .

limn→∞

(−1)n+1 n

Alternating Series

n=1(−1)nbn orP∞

I Example∞Xn=1

(−1)n

n= −1 +

2− 1

4− 1

5+ . . .

∞Xn=1

(−1)n+1 n

2n + 1=

3− 2

7− 4

9+ . . .

limn→∞

(−1)n+1 n

Alternating Series test

We have the following test for such alternating series:Alternating Series test If the alternating series

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . bn > 0

satisfies(i) bn+1 ≤ bn for all n

(ii) limn→∞

bn = 0

then the series converges.

I we see from the graph that because the values of bn are decreasing, thepartial sums of the series cluster about some point in the interval [0, b1].

I A proof is given at the end of the notes.

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . bn > 0

(ii) limn→∞

bn = 0

I A proof is given at the end of the notes.

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . bn > 0

(ii) limn→∞

bn = 0

I A proof is given at the end of the notes.Annette Pilkington Lecture 27 :Alternating Series

Alternating Series test If the alternating series

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . , bn > 0

(ii) limn→∞

bn = 0

I A similar theorem applies to the seriesP∞

i=1(−1)nbn.

I Also we really only need bn+1 ≤ bn for all n > N for some N, since a finitenumber of terms do not change whether a series converges or not.

I Recall that if we have a differentiable function f (x), with f (n) = bn, thenwe can use its derivative to check if terms are decreasing.

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . , bn > 0

(ii) limn→∞

bn = 0

i=1(−1)nbn.

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . , bn > 0

(ii) limn→∞

bn = 0

i=1(−1)nbn.

∞Xn=1

(−1)n−1bn = b1 − b2 + b3 − b4 + . . . , bn > 0

(ii) limn→∞

bn = 0

i=1(−1)nbn.

Example 1

Alternating Series test If the alternating seriesP∞n=1(−1)n−1bn = b1 − b2 + b3 − b4 + . . . bn > 0 satisfies

(i) bn+1 ≤ bn for all n

(ii) limn→∞

bn = 0

then the series converges.Example 1 Test the following series for convergence

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

I Therefore, we can conclude that the alternating seriesP∞

n=1(−1)n 1n

converges.

I Note that an alternating series may converge whilst the sum of theabsolute values diverges. In particular the alternating harmonic seriesabove converges.

Example 1

(ii) limn→∞

bn = 0

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

n=1(−1)n 1n

converges.

Example 1

(ii) limn→∞

bn = 0

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

n=1(−1)n 1n

converges.

Example 1

(ii) limn→∞

bn = 0

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

n=1(−1)n 1n

converges.

Example 1

(ii) limn→∞

bn = 0

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

n=1(−1)n 1n

converges.

Example 1

(ii) limn→∞

bn = 0

∞Xn=1

(−1)n 1

I We have bn = 1n

I limn→∞1n

I bn+1 = 1n+1

< bn = 1n

for all n ≥ 1.

n=1(−1)n 1n

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I We have bn = nn2+1

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I To check if the terms bn decrease as n increases, we use a derivative. Letf (x) = x

x2+1. We have f (n) = bn.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

I Since this function is decreasing as x increases, for x > 1, we must havebn+1 < bn for n ≥ 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 2

(ii) limn→∞

bn = 0

P∞n=1(−1)n n

I limn→∞n

n2+1= limn→∞

1+1/n2 = 0.

I f ′(x) = (x2+1)−x(2x)

(x2+1)2= 1−x2

(x2+1)2< 0 for x > 1.

n=1(−1)n nn2+1

converges.

Example 3

(ii) limn→∞

bn = 0

Example 3 Test the following series for convergence:P∞

n=1(−1)n 2n2

I We have bn = 2n2

I Here we can use the divergence test (you should always check if thisapplies first)

I We have limn→∞2n2

n2+1= limn→∞

21+1/n2 = 2 6= 0.

I Therefore limn→∞(−1)n 2n2

n2+1does not exist and we can conclude that the

series∞Xn=1

(−1)n 2n2

n2 + 1

diverges.

Example 3

(ii) limn→∞

bn = 0

n=1(−1)n 2n2

I We have bn = 2n2

n2+1= limn→∞

21+1/n2 = 2 6= 0.

series∞Xn=1

(−1)n 2n2

n2 + 1

diverges.

Example 3

(ii) limn→∞

bn = 0

n=1(−1)n 2n2

I We have bn = 2n2

n2+1= limn→∞

21+1/n2 = 2 6= 0.

series∞Xn=1

(−1)n 2n2

n2 + 1

diverges.

Example 3

(ii) limn→∞

bn = 0

n=1(−1)n 2n2

I We have bn = 2n2

n2+1= limn→∞

21+1/n2 = 2 6= 0.

series∞Xn=1

(−1)n 2n2

n2 + 1

diverges.

Example 3

(ii) limn→∞

bn = 0

n=1(−1)n 2n2

I We have bn = 2n2

n2+1= limn→∞

21+1/n2 = 2 6= 0.

series∞Xn=1

(−1)n 2n2

n2 + 1

diverges.

Example 4

(ii) limn→∞

bn = 0

then the series converges.Example 4 Test the following series for convergence:

P∞n=1(−1)n 1

I We have bn = 1n!

I Since 0 ≤ bn = 1n·(n−1)·(n−2)·····1 ≤

, we must have limn→∞1n!

I bn+1 = 1(n+1)·n·(n−1)·(n−2)·····1 = 1

(n+1)· 1

n·(n−1)·(n−2)·····1 = 1n+1· bn < bn if

n > 1.

I Therefore by the Alternating series test, we can conclude that the seriesP∞n=1(−1)n 1

n!converges.

Example 4

(ii) limn→∞

bn = 0

P∞n=1(−1)n 1

I We have bn = 1n!

I Since 0 ≤ bn = 1n·(n−1)·(n−2)·····1 ≤

I bn+1 = 1(n+1)·n·(n−1)·(n−2)·····1 = 1

(n+1)· 1

n·(n−1)·(n−2)·····1 = 1n+1· bn < bn if

n > 1.

n!converges.

Example 4

(ii) limn→∞

bn = 0

P∞n=1(−1)n 1

I We have bn = 1n!

I Since 0 ≤ bn = 1n·(n−1)·(n−2)·····1 ≤

I bn+1 = 1(n+1)·n·(n−1)·(n−2)·····1 = 1

(n+1)· 1

n·(n−1)·(n−2)·····1 = 1n+1· bn < bn if

n > 1.

n!converges.

Example 4

(ii) limn→∞

bn = 0

P∞n=1(−1)n 1

I We have bn = 1n!

I Since 0 ≤ bn = 1n·(n−1)·(n−2)·····1 ≤

I bn+1 = 1(n+1)·n·(n−1)·(n−2)·····1 = 1

(n+1)· 1

n·(n−1)·(n−2)·····1 = 1n+1· bn < bn if

n > 1.

n!converges.

Example 4

(ii) limn→∞

bn = 0

P∞n=1(−1)n 1

I We have bn = 1n!

I Since 0 ≤ bn = 1n·(n−1)·(n−2)·····1 ≤

I bn+1 = 1(n+1)·n·(n−1)·(n−2)·····1 = 1

(n+1)· 1

n·(n−1)·(n−2)·····1 = 1n+1· bn < bn if

n > 1.

n!converges.

Example 5

n=1(−1)n ln nn2

I We have bn = ln nn2 .

I limn→∞ln nn2 = limx→∞

ln xx2 = (L′Hop) limx→∞

= limx→∞1

2x2 = 0.

I To check if bn is decreasing as n increases, we calculate the derivative off (x) = ln x

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

x2 < 0 if 1− 2 ln x < 0 orln x > 1/2. This happens if x >

√e, which certainly happens if x ≥ 2.

I This is enough to show that bn+1 < bn if n ≥ 2 and henceP∞

n=1(−1)n ln nn2

converges.

Example 5

n=1(−1)n ln nn2

= limx→∞1

2x2 = 0.

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

n=1(−1)n ln nn2

converges.

Example 5

n=1(−1)n ln nn2

= limx→∞1

2x2 = 0.

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

n=1(−1)n ln nn2

converges.

Example 5

n=1(−1)n ln nn2

= limx→∞1

2x2 = 0.

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

n=1(−1)n ln nn2

converges.

Example 5

n=1(−1)n ln nn2

= limx→∞1

2x2 = 0.

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

n=1(−1)n ln nn2

converges.

Example 5

n=1(−1)n ln nn2

= limx→∞1

2x2 = 0.

I f ′(x) = (x2)(1/x)−2x ln x

x2 = x−2x ln xx2 = x(1−2 ln x)

n=1(−1)n ln nn2

converges.

Example 6

n=1(−1)n cos“πn

I We have bn = cos“πn

”. bn > 0 for n ≥ 2.

I limn→∞ cos“πn

”= limx→∞ cos

“πx

”= 1 6= 0.

I Therefore limn→∞(−1)n cos“πn

”does not exist and the seriesP∞

”diverges by the divergence test.

Example 6

”I We have bn = cos

“πn

”. bn > 0 for n ≥ 2.

”= limx→∞ cos

“πx

”= 1 6= 0.

Example 6

“πn

”. bn > 0 for n ≥ 2.

”= limx→∞ cos

“πx

”= 1 6= 0.

Example 6

“πn

”. bn > 0 for n ≥ 2.

”= limx→∞ cos

“πx

”= 1 6= 0.

Error of Estimation

Estimating the Error

SupposeP∞

i=1(−1)n−1bn, bn > 0, converges to s. Recall that we can use thepartial sum sn = b1 − b2 + · · ·+ (−1)n−1bn to estimate the sum of the series, s.If the series satisfies the conditions for the Alternating series test, we have thefollowing simple estimate of the size of the error in our approximation|Rn| = |s − sn|.(Rn here stands for the remainder when we subtract the n th partial sum fromthe sum of the series. )

Alternating Series Estimation Theorem If s =P

(−1)n−1bn, bn > 0 is thesum of an alternating series that satisfies

(i) bn+1 < bn for all n

(ii) limn→∞

bn = 0

then|Rn| = |s − sn| ≤ bn+1.

A proof is included at the end of the notes.

Example

(ii) limn→∞

bn = 0

Example Find a partial sum approximation the sum of the seriesP∞

n=1(−1)n 1n

where the error of approximation is less than .01 = 10−2.

I We have bn = 1n

. bn > 0 for n ≥ 1 and we have already seen that theconditions of the alternating series test are satisfied in a previous example.

I Therefore the n th remainder, |Rn| = |s − sn| ≤ bn+1 = 1n+1

I Therefore, if we find a value of n for which 1n+1≤ 1

102 , we will have the

error of approximation Rn ≤ 1102 .

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

I Checking with Mathematica, we get the actual errorR101 =

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

= 0.00492599 which is indeed lessthan .01.

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Example

(ii) limn→∞

bn = 0

n=1(−1)n 1n

I We have bn = 1n

I 1n+1≤ 1

102 if 102 ≤ n + 1 or n ≥ 101.

P∞n=1(−1)n 1

n−P101

n=1(−1)n 1n

Lecture 27 :Alternating Series - University of Notre Dameapilking/Math10560/Lectures/Lecture...

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