Lecture 27 :Alternating Series I Example ∞ X (−1)N 1 1 1 1 = −1 + − + − +

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P∞ n P∞ n+1 I A series of the form n=1(−1) bn or n=1(−1) bn, where bn > 0 for all n, is called an alternating series, because the terms alternate between positive and negative values.

I Example ∞ X (−1)n 1 1 1 1 = −1 + − + − + ... n 2 3 4 5 n=1 ∞ X n+1 n 1 2 3 4 (−1) = − + − + ... 2n + 1 3 5 7 9 n=1

I We can use the divergence test to show that the second series above diverges, since n lim (−1)n+1 does not exist n→∞ 2n + 1

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 only to series with positive terms.

Annette Pilkington Lecture 27 :Alternating Series I Example ∞ X (−1)n 1 1 1 1 = −1 + − + − + ... n 2 3 4 5 n=1 ∞ X n+1 n 1 2 3 4 (−1) = − + − + ... 2n + 1 3 5 7 9 n=1

I We can use the divergence test to show that the second series above diverges, since n lim (−1)n+1 does not exist n→∞ 2n + 1

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 only to series with positive terms. P∞ n P∞ n+1 I A series of the form n=1(−1) bn or n=1(−1) bn, where bn > 0 for all n, is called an alternating series, because the terms alternate between positive and negative values.

Annette Pilkington Lecture 27 :Alternating Series I We can use the divergence test to show that the second series above diverges, since n lim (−1)n+1 does not exist n→∞ 2n + 1

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

I Example ∞ X (−1)n 1 1 1 1 = −1 + − + − + ... n 2 3 4 5 n=1 ∞ X n+1 n 1 2 3 4 (−1) = − + − + ... 2n + 1 3 5 7 9 n=1

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Alternating Series

I Example ∞ X (−1)n 1 1 1 1 = −1 + − + − + ... n 2 3 4 5 n=1 ∞ X n+1 n 1 2 3 4 (−1) = − + − + ... 2n + 1 3 5 7 9 n=1

I We can use the divergence test to show that the second series above diverges, since n lim (−1)n+1 does not exist n→∞ 2n + 1

Annette Pilkington Lecture 27 :Alternating Series I we see from the graph that because the values of bn are decreasing, the partial sums of the series cluster about some point in the interval [0, b1].

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

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

We have the following test for such alternating series: Alternating Series test If the alternating series ∞ X n−1 (−1) bn = b1 − b2 + b3 − b4 + ... bn > 0 n=1 satisﬁes (i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges.

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

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

(ii) lim bn = 0 n→∞ then the series converges.

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

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Alternating Series test

(ii) lim bn = 0 n→∞ then the series converges.

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

I A proof is given at the end of the notes. Annette Pilkington Lecture 27 :Alternating Series P∞ n I A similar theorem applies to the series i=1(−1) bn.

I Also we really only need bn+1 ≤ bn for all n > N for some N, since a ﬁnite number 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, then we can use its derivative to check if terms are decreasing.

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

Alternating Series test If the alternating series

∞ X n−1 (−1) bn = b1 − b2 + b3 − b4 + ..., bn > 0 n=1 satisﬁes (i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges.

Annette Pilkington Lecture 27 :Alternating Series I Also we really only need bn+1 ≤ bn for all n > N for some N, since a ﬁnite number 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, then we can use its derivative to check if terms are decreasing.

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

Alternating Series test If the alternating series

∞ X n−1 (−1) bn = b1 − b2 + b3 − b4 + ..., bn > 0 n=1 satisﬁes (i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n I A similar theorem applies to the series i=1(−1) bn.

Annette Pilkington Lecture 27 :Alternating Series I Recall that if we have a diﬀerentiable function f (x), with f (n) = bn, then we can use its derivative to check if terms are decreasing.

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

Alternating Series test If the alternating series

∞ X n−1 (−1) bn = b1 − b2 + b3 − b4 + ..., bn > 0 n=1 satisﬁes (i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n I A similar theorem applies to the series i=1(−1) bn.

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

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Notes

Alternating Series test If the alternating series

∞ X n−1 (−1) bn = b1 − b2 + b3 − b4 + ..., bn > 0 n=1 satisﬁes (i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n I A similar theorem applies to the series i=1(−1) bn.

I Also we really only need bn+1 ≤ bn for all n > N for some N, since a ﬁnite number 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, then we can use its derivative to check if terms are decreasing.

Annette Pilkington Lecture 27 :Alternating Series 1 I We have bn = n . 1 I limn→∞ n = 0. 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1. P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

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

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

Annette Pilkington Lecture 27 :Alternating Series 1 I limn→∞ n = 0. 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1. P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

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

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

1 I We have bn = n .

Annette Pilkington Lecture 27 :Alternating Series 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1. P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

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

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

1 I We have bn = n . 1 I limn→∞ n = 0.

Annette Pilkington Lecture 27 :Alternating Series P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

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

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

1 I We have bn = n . 1 I limn→∞ n = 0. 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1.

Annette Pilkington Lecture 27 :Alternating Series I Note that an alternating series may converge whilst the sum of the absolute values diverges. In particular the alternating harmonic series above converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

1 I We have bn = n . 1 I limn→∞ n = 0. 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1. P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 1

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. Example 1 Test the following series for convergence ∞ X n 1 (−1) n n=1

1 I We have bn = n . 1 I limn→∞ n = 0. 1 1 I bn+1 = n+1 < bn = n for all n ≥ 1. P∞ n 1 I Therefore, we can conclude that the alternating series n=1(−1) n converges.

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

Annette Pilkington Lecture 27 :Alternating Series n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

I To check if the terms bn decrease as n increases, we use a derivative. Let x f (x) = x2+1 . We have f (n) = bn. 0 (x2+1)−x(2x) 1−x2 I f (x) = (x2+1)2 = (x2+1)2 < 0 for x > 1. I Since this function is decreasing as x increases, for x > 1, we must have bn+1 < bn for n ≥ 1. P∞ n n I Therefore, we can conclude that the alternating series n=1(−1) n2+1 converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1

Annette Pilkington Lecture 27 :Alternating Series n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 .

Annette Pilkington Lecture 27 :Alternating Series I To check if the terms bn decrease as n increases, we use a derivative. Let x f (x) = x2+1 . We have f (n) = bn. 0 (x2+1)−x(2x) 1−x2 I f (x) = (x2+1)2 = (x2+1)2 < 0 for x > 1. I Since this function is decreasing as x increases, for x > 1, we must have bn+1 < bn for n ≥ 1. P∞ n n I Therefore, we can conclude that the alternating series n=1(−1) n2+1 converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

Annette Pilkington Lecture 27 :Alternating Series 0 (x2+1)−x(2x) 1−x2 I f (x) = (x2+1)2 = (x2+1)2 < 0 for x > 1. I Since this function is decreasing as x increases, for x > 1, we must have bn+1 < bn for n ≥ 1. P∞ n n I Therefore, we can conclude that the alternating series n=1(−1) n2+1 converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

I To check if the terms bn decrease as n increases, we use a derivative. Let x f (x) = x2+1 . We have f (n) = bn.

Annette Pilkington Lecture 27 :Alternating Series I Since this function is decreasing as x increases, for x > 1, we must have bn+1 < bn for n ≥ 1. P∞ n n I Therefore, we can conclude that the alternating series n=1(−1) n2+1 converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

I To check if the terms bn decrease as n increases, we use a derivative. Let x f (x) = x2+1 . We have f (n) = bn. 0 (x2+1)−x(2x) 1−x2 I f (x) = (x2+1)2 = (x2+1)2 < 0 for x > 1.

Annette Pilkington Lecture 27 :Alternating Series P∞ n n I Therefore, we can conclude that the alternating series n=1(−1) n2+1 converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 2

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n n Example 2 Test the following series for convergence n=1(−1) n2+1 n I We have bn = n2+1 . n 1/n I limn→∞ n2+1 = limn→∞ 1+1/n2 = 0.

Annette Pilkington Lecture 27 :Alternating Series 2n2 I We have bn = n2+1 . I Here we can use the divergence test (you should always check if this applies ﬁrst)

2n2 2 I We have limn→∞ n2+1 = limn→∞ 1+1/n2 = 2 6= 0. n 2n2 I Therefore limn→∞(−1) n2+1 does not exist and we can conclude that the series ∞ 2 X n 2n (−1) n2 + 1 n=1 diverges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 2n2 Example 3 Test the following series for convergence: n=1(−1) n2+1

Annette Pilkington Lecture 27 :Alternating Series I Here we can use the divergence test (you should always check if this applies ﬁrst)

2n2 2 I We have limn→∞ n2+1 = limn→∞ 1+1/n2 = 2 6= 0. n 2n2 I Therefore limn→∞(−1) n2+1 does not exist and we can conclude that the series ∞ 2 X n 2n (−1) n2 + 1 n=1 diverges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 2n2 Example 3 Test the following series for convergence: n=1(−1) n2+1 2n2 I We have bn = n2+1 .

Annette Pilkington Lecture 27 :Alternating Series 2n2 2 I We have limn→∞ n2+1 = limn→∞ 1+1/n2 = 2 6= 0. n 2n2 I Therefore limn→∞(−1) n2+1 does not exist and we can conclude that the series ∞ 2 X n 2n (−1) n2 + 1 n=1 diverges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 2n2 Example 3 Test the following series for convergence: n=1(−1) n2+1 2n2 I We have bn = n2+1 . I Here we can use the divergence test (you should always check if this applies ﬁrst)

Annette Pilkington Lecture 27 :Alternating Series n 2n2 I Therefore limn→∞(−1) n2+1 does not exist and we can conclude that the series ∞ 2 X n 2n (−1) n2 + 1 n=1 diverges.

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

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

(i) bn+1 ≤ bn for all n

2n2 2 I We have limn→∞ n2+1 = limn→∞ 1+1/n2 = 2 6= 0.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 3

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

(i) bn+1 ≤ bn for all n

2n2 2 I We have limn→∞ n2+1 = limn→∞ 1+1/n2 = 2 6= 0. n 2n2 I Therefore limn→∞(−1) n2+1 does not exist and we can conclude that the series ∞ 2 X n 2n (−1) n2 + 1 n=1 diverges.

Annette Pilkington Lecture 27 :Alternating Series 1 I We have bn = n! . 1 1 1 I Since 0 ≤ bn = n·(n−1)·(n−2)·····1 ≤ n , we must have limn→∞ n! = 0. 1 1 1 1 I bn+1 = (n+1)·n·(n−1)·(n−2)·····1 = (n+1) · n·(n−1)·(n−2)·····1 = n+1 · bn < bn if n > 1.

I Therefore by the Alternating series test, we can conclude that the series P∞ n 1 n=1(−1) n! converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 1 Example 4 Test the following series for convergence: n=1(−1) n!

Annette Pilkington Lecture 27 :Alternating Series 1 1 1 I Since 0 ≤ bn = n·(n−1)·(n−2)·····1 ≤ n , we must have limn→∞ n! = 0. 1 1 1 1 I bn+1 = (n+1)·n·(n−1)·(n−2)·····1 = (n+1) · n·(n−1)·(n−2)·····1 = n+1 · bn < bn if n > 1.

I Therefore by the Alternating series test, we can conclude that the series P∞ n 1 n=1(−1) n! converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 1 Example 4 Test the following series for convergence: n=1(−1) n! 1 I We have bn = n! .

Annette Pilkington Lecture 27 :Alternating Series 1 1 1 1 I bn+1 = (n+1)·n·(n−1)·(n−2)·····1 = (n+1) · n·(n−1)·(n−2)·····1 = n+1 · bn < bn if n > 1.

I Therefore by the Alternating series test, we can conclude that the series P∞ n 1 n=1(−1) n! converges.

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

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

(i) bn+1 ≤ bn for all n

Annette Pilkington Lecture 27 :Alternating Series I Therefore by the Alternating series test, we can conclude that the series P∞ n 1 n=1(−1) n! converges.

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

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

(i) bn+1 ≤ bn for all n

(ii) lim bn = 0 n→∞ then the series converges. P∞ n 1 Example 4 Test the following series for convergence: n=1(−1) n! 1 I We have bn = n! . 1 1 1 I Since 0 ≤ bn = n·(n−1)·(n−2)·····1 ≤ n , we must have limn→∞ n! = 0. 1 1 1 1 I bn+1 = (n+1)·n·(n−1)·(n−2)·····1 = (n+1) · n·(n−1)·(n−2)·····1 = n+1 · bn < bn if n > 1.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 4

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

(i) bn+1 ≤ bn for all n

I Therefore by the Alternating series test, we can conclude that the series P∞ n 1 n=1(−1) n! converges.

Annette Pilkington Lecture 27 :Alternating Series ln n I We have bn = n2 . ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

I To check if bn is decreasing as n increases, we calculate the derivative of ln x f (x) = x2 . 2 f 0(x) = (x )(1/x)−2x ln x = x−2x ln x = x(1−2 ln x) < 0 if 1 − 2 ln x < 0 or I x2 x2 √ x2 ln x > 1/2. This happens if x > e, which certainly happens if x ≥ 2. P∞ n ln n I This is enough to show that bn+1 < bn if n ≥ 2 and hence n=1(−1) n2 converges.

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

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2

Annette Pilkington Lecture 27 :Alternating Series ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

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

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2 ln n I We have bn = n2 .

Annette Pilkington Lecture 27 :Alternating Series I To check if bn is decreasing as n increases, we calculate the derivative of ln x f (x) = x2 . 2 f 0(x) = (x )(1/x)−2x ln x = x−2x ln x = x(1−2 ln x) < 0 if 1 − 2 ln x < 0 or I x2 x2 √ x2 ln x > 1/2. This happens if x > e, which certainly happens if x ≥ 2. P∞ n ln n I This is enough to show that bn+1 < bn if n ≥ 2 and hence n=1(−1) n2 converges.

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

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2 ln n I We have bn = n2 . ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

Annette Pilkington Lecture 27 :Alternating Series 2 f 0(x) = (x )(1/x)−2x ln x = x−2x ln x = x(1−2 ln x) < 0 if 1 − 2 ln x < 0 or I x2 x2 √ x2 ln x > 1/2. This happens if x > e, which certainly happens if x ≥ 2. P∞ n ln n I This is enough to show that bn+1 < bn if n ≥ 2 and hence n=1(−1) n2 converges.

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

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2 ln n I We have bn = n2 . ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

I To check if bn is decreasing as n increases, we calculate the derivative of ln x f (x) = x2 .

Annette Pilkington Lecture 27 :Alternating Series P∞ n ln n I This is enough to show that bn+1 < bn if n ≥ 2 and hence n=1(−1) n2 converges.

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

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2 ln n I We have bn = n2 . ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 5

P∞ n ln n Example 5 Test the following series for convergence: n=1(−1) n2 ln n I We have bn = n2 . ln n ln x 0 1/x 1 I limn→∞ n2 = limx→∞ x2 =( L Hop) limx→∞ 2x = limx→∞ 2x2 = 0.

Annette Pilkington Lecture 27 :Alternating Series “ π ” I We have bn = cos n . bn > 0 for n ≥ 2.

“ π ” “ π ” I limn→∞ cos n = limx→∞ cos x = 1 6= 0.

n “ π ” I Therefore limn→∞(−1) cos n does not exist and the series P∞ n “ π ” n=1(−1) cos n diverges by the divergence test.

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

P∞ n “ π ” Example 6 Test the following series for convergence: n=1(−1) cos n

Annette Pilkington Lecture 27 :Alternating Series “ π ” “ π ” I limn→∞ cos n = limx→∞ cos x = 1 6= 0.

n “ π ” I Therefore limn→∞(−1) cos n does not exist and the series P∞ n “ π ” n=1(−1) cos n diverges by the divergence test.

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

P∞ n “ π ” Example 6 Test the following series for convergence: n=1(−1) cos n

“ π ” I We have bn = cos n . bn > 0 for n ≥ 2.

Annette Pilkington Lecture 27 :Alternating Series n “ π ” I Therefore limn→∞(−1) cos n does not exist and the series P∞ n “ π ” n=1(−1) cos n diverges by the divergence test.

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

P∞ n “ π ” Example 6 Test the following series for convergence: n=1(−1) cos n

“ π ” I We have bn = cos n . bn > 0 for n ≥ 2.

“ π ” “ π ” I limn→∞ cos n = limx→∞ cos x = 1 6= 0.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example 6

P∞ n “ π ” Example 6 Test the following series for convergence: n=1(−1) cos n

“ π ” I We have bn = cos n . bn > 0 for n ≥ 2.

“ π ” “ π ” I limn→∞ cos n = limx→∞ cos x = 1 6= 0.

n “ π ” I Therefore limn→∞(−1) cos n does not exist and the series P∞ n “ π ” n=1(−1) cos n diverges by the divergence test.

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Error of Estimation

Estimating the Error

P∞ n−1 Suppose i=1(−1) bn, bn > 0, converges to s. Recall that we can use the n−1 partial sum sn = b1 − b2 + ··· + (−1) bn to estimate the sum of the series, s. If the series satisﬁes the conditions for the Alternating series test, we have the following 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 from the sum of the series. ) P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

(ii) lim bn = 0 n→∞ then |Rn| = |s − sn| ≤ bn+1. A proof is included at the end of the notes.

Annette Pilkington Lecture 27 :Alternating Series 1 I We have bn = n . bn > 0 for n ≥ 1 and we have already seen that the conditions of the alternating series test are satisﬁed in a previous example. 1 I Therefore the n th remainder, |Rn| = |s − sn| ≤ bn+1 = n+1 . 1 1 I Therefore, if we ﬁnd a value of n for which n+1 ≤ 102 , we will have the 1 error of approximation Rn ≤ 102 . 1 1 2 I n+1 ≤ 102 if 10 ≤ n + 1 or n ≥ 101. I Checking with Mathematica, we get the actual error P∞ n 1 P101 n 1 R101 = n=1(−1) n − n=1(−1) n = 0.00492599 which is indeed less than .01.

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

(ii) lim bn = 0 n→∞ then |Rn| = |s − sn| ≤ bn+1. P∞ n 1 Example Find a partial sum approximation the sum of the series n=1(−1) n where the error of approximation is less than .01 = 10−2.

Annette Pilkington Lecture 27 :Alternating Series 1 I Therefore the n th remainder, |Rn| = |s − sn| ≤ bn+1 = n+1 . 1 1 I Therefore, if we ﬁnd a value of n for which n+1 ≤ 102 , we will have the 1 error of approximation Rn ≤ 102 . 1 1 2 I n+1 ≤ 102 if 10 ≤ n + 1 or n ≥ 101. I Checking with Mathematica, we get the actual error P∞ n 1 P101 n 1 R101 = n=1(−1) n − n=1(−1) n = 0.00492599 which is indeed less than .01.

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

Annette Pilkington Lecture 27 :Alternating Series 1 1 I Therefore, if we ﬁnd a value of n for which n+1 ≤ 102 , we will have the 1 error of approximation Rn ≤ 102 . 1 1 2 I n+1 ≤ 102 if 10 ≤ n + 1 or n ≥ 101. I Checking with Mathematica, we get the actual error P∞ n 1 P101 n 1 R101 = n=1(−1) n − n=1(−1) n = 0.00492599 which is indeed less than .01.

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

Annette Pilkington Lecture 27 :Alternating Series 1 1 2 I n+1 ≤ 102 if 10 ≤ n + 1 or n ≥ 101. I Checking with Mathematica, we get the actual error P∞ n 1 P101 n 1 R101 = n=1(−1) n − n=1(−1) n = 0.00492599 which is indeed less than .01.

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

(ii) lim bn = 0 n→∞ then |Rn| = |s − sn| ≤ bn+1. P∞ n 1 Example Find a partial sum approximation the sum of the series n=1(−1) n where the error of approximation is less than .01 = 10−2. 1 I We have bn = n . bn > 0 for n ≥ 1 and we have already seen that the conditions of the alternating series test are satisﬁed in a previous example. 1 I Therefore the n th remainder, |Rn| = |s − sn| ≤ bn+1 = n+1 . 1 1 I Therefore, if we ﬁnd a value of n for which n+1 ≤ 102 , we will have the 1 error of approximation Rn ≤ 102 .

Annette Pilkington Lecture 27 :Alternating Series I Checking with Mathematica, we get the actual error P∞ n 1 P101 n 1 R101 = n=1(−1) n − n=1(−1) n = 0.00492599 which is indeed less than .01.

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

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

Annette Pilkington Lecture 27 :Alternating Series Alternating Series Alternating Series test Notes Example 1 Example 2 Example 3 Example 4 Example 5 Example 6 Error of Estimation Example

P n−1 Alternating Series Estimation Theorem If s = (−1) bn, bn > 0 is the sum of an alternating series that satisﬁes

(i) bn+1 < bn for all n

Annette Pilkington Lecture 27 :Alternating Series