Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example P-Series Integral Test

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I Note The result is still true if the condition that f (x) is decreasing on the interval [1, ∞) is relaxed to “the function f (x) is decreasing on an interval [M, ∞) for some number M ≥ 1.”

Integral Test Integral Test Example Integral Test Example p-series Integral Test

In this section, we see that we can sometimes decide whether a series converges or diverges by comparing it to an improper integral. The analysis in this section P only applies to series an, with positive terms, that is an > 0. Integral Test Suppose f (x) is a positive decreasing continuous function on the interval [1, ∞) with f (n) = an. P∞ R ∞ Then the series n=1 an is convergent if and only if 1 f (x)dx converges, that is: ∞ Z ∞ X If f (x)dx is convergent, then an is convergent. 1 n=1 ∞ Z ∞ X If f (x)dx is divergent, then an is divergent. 1 n=1

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral Test

Annette Pilkington Lecture 25 : Integral Test I In the picture we compare the P∞ 1 series n=1 n2 to the improper R ∞ 1 integral 1 x2 dx.

Pn 1 R ∞ 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2.

I Since the sequence {sn} is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series ∞ X 1 converges. n2 i=1 P∞ 1 R ∞ 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence)

We know from a previous lecture that R ∞ 1 1 xp dx converges if p > 1 and diverges if p ≤ 1.

Annette Pilkington Lecture 25 : Integral Test Pn 1 R ∞ 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence)

We know from a previous lecture that R ∞ 1 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture we compare the P∞ 1 series n=1 n2 to the improper R ∞ 1 integral 1 x2 dx.

Annette Pilkington Lecture 25 : Integral Test I Since the sequence {sn} is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series ∞ X 1 converges. n2 i=1 P∞ 1 R ∞ 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence)

We know from a previous lecture that R ∞ 1 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture we compare the P∞ 1 series n=1 n2 to the improper R ∞ 1 integral 1 x2 dx.

Pn 1 R ∞ 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2.

Annette Pilkington Lecture 25 : Integral Test P∞ 1 R ∞ 1 I NOTE We are not saying that i=1 n2 = 1 x2 dx here.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence)

We know from a previous lecture that R ∞ 1 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture we compare the P∞ 1 series n=1 n2 to the improper R ∞ 1 integral 1 x2 dx.

Pn 1 R ∞ 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2.

I Since the sequence {sn} is increasing (because each an > 0) and bounded, we can conclude that the sequence of partial sums converges and hence the series ∞ X 1 converges. n2 i=1

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: convergence)

We know from a previous lecture that R ∞ 1 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture we compare the P∞ 1 series n=1 n2 to the improper R ∞ 1 integral 1 x2 dx.

Pn 1 R ∞ 1 I The n th partial sum is sn = 1 + n=2 n2 < 1 + 1 x2 dx = 1 + 1 = 2.

Annette Pilkington Lecture 25 : Integral Test I In the picture, we compare the ∞ P √1 series n=1 n to the improper ∞ R √1 integral 1 x dx.

∞ X 1 1 1 1 √ = √ + √ + √ + ··· k=1 n 1 2 3

I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = √ + √ + √ + ··· + √ > √ dx 1 2 3 n − 1 1 x

n R √1 I Thus we see that limn→∞ sn > limn→∞ 1 x dx. n R √1 I However, we know that 1 x dx grows without bound and hence since ∞ ∞ R √1 P √1 1 x dx diverges, we can conclude that k=1 n also diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence)

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

Annette Pilkington Lecture 25 : Integral Test I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = √ + √ + √ + ··· + √ > √ dx 1 2 3 n − 1 1 x

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence)

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture, we compare the ∞ P √1 series n=1 n to the improper ∞ R √1 integral 1 x dx.

∞ X 1 1 1 1 √ = √ + √ + √ + ··· k=1 n 1 2 3

Annette Pilkington Lecture 25 : Integral Test n R √1 I Thus we see that limn→∞ sn > limn→∞ 1 x dx. n R √1 I However, we know that 1 x dx grows without bound and hence since ∞ ∞ R √1 P √1 1 x dx diverges, we can conclude that k=1 n also diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence)

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture, we compare the ∞ P √1 series n=1 n to the improper ∞ R √1 integral 1 x dx.

∞ X 1 1 1 1 √ = √ + √ + √ + ··· k=1 n 1 2 3

I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = √ + √ + √ + ··· + √ > √ dx 1 2 3 n − 1 1 x

Annette Pilkington Lecture 25 : Integral Test n R √1 I However, we know that 1 x dx grows without bound and hence since ∞ ∞ R √1 P √1 1 x dx diverges, we can conclude that k=1 n also diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence)

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture, we compare the ∞ P √1 series n=1 n to the improper ∞ R √1 integral 1 x dx.

∞ X 1 1 1 1 √ = √ + √ + √ + ··· k=1 n 1 2 3

I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = √ + √ + √ + ··· + √ > √ dx 1 2 3 n − 1 1 x

n R √1 I Thus we see that limn→∞ sn > limn→∞ 1 x dx.

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral Test (Why it works: divergence)

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

I In the picture, we compare the ∞ P √1 series n=1 n to the improper ∞ R √1 integral 1 x dx.

∞ X 1 1 1 1 √ = √ + √ + √ + ··· k=1 n 1 2 3

I This time we draw the rectangles so that we get 1 1 1 1 Z n 1 sn > sn−1 = √ + √ + √ + ··· + √ > √ dx 1 2 3 n − 1 1 x

Annette Pilkington Lecture 25 : Integral Test 2 I Consider the function f (x) = 3x+5 . I This function is positive and continuous on the interval [1, ∞). We see 0 −6 that it is decreasing by examining the derivative. f (x) = (3x+5)2 < 0. 2 I Since f (n) = 3n+5 , to determine convergence or divergence of the series P∞ 2 R ∞ n=1 3n+5 it suﬃces to check what happens to 1 f (x)dx. R ∞ 2 R t 2 R 3t+5 2 I 1 3x+5 dx = limt→∞ 1 3x+5 dx = 7 3u du(u = 3x + 5) = ˛3t+5 2 ˛ 2 limt→∞ 3 ln |u| = limt→∞ 3 [ln |3t + 5| − ln |7|] = ∞. (integral ˛7 diverges). P∞ 2 I Therefore, the series n=1 3n+5 diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Integral Test Suppose f (x) is a positive decreasing continuous function on P∞ the interval [1, ∞) with f (n) = an. Then the series n=1 an is convergent if R ∞ and only if 1 f (x)dx converges Example Use the integral test to determine if the following series converges:

∞ X 2 3n + 5 n=1

Annette Pilkington Lecture 25 : Integral Test I This function is positive and continuous on the interval [1, ∞). We see 0 −6 that it is decreasing by examining the derivative. f (x) = (3x+5)2 < 0. 2 I Since f (n) = 3n+5 , to determine convergence or divergence of the series P∞ 2 R ∞ n=1 3n+5 it suﬃces to check what happens to 1 f (x)dx. R ∞ 2 R t 2 R 3t+5 2 I 1 3x+5 dx = limt→∞ 1 3x+5 dx = 7 3u du(u = 3x + 5) = ˛3t+5 2 ˛ 2 limt→∞ 3 ln |u| = limt→∞ 3 [ln |3t + 5| − ln |7|] = ∞. (integral ˛7 diverges). P∞ 2 I Therefore, the series n=1 3n+5 diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

∞ X 2 3n + 5 n=1

2 I Consider the function f (x) = 3x+5 .

Annette Pilkington Lecture 25 : Integral Test 2 I Since f (n) = 3n+5 , to determine convergence or divergence of the series P∞ 2 R ∞ n=1 3n+5 it suﬃces to check what happens to 1 f (x)dx. R ∞ 2 R t 2 R 3t+5 2 I 1 3x+5 dx = limt→∞ 1 3x+5 dx = 7 3u du(u = 3x + 5) = ˛3t+5 2 ˛ 2 limt→∞ 3 ln |u| = limt→∞ 3 [ln |3t + 5| − ln |7|] = ∞. (integral ˛7 diverges). P∞ 2 I Therefore, the series n=1 3n+5 diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

∞ X 2 3n + 5 n=1

2 I Consider the function f (x) = 3x+5 . I This function is positive and continuous on the interval [1, ∞). We see 0 −6 that it is decreasing by examining the derivative. f (x) = (3x+5)2 < 0.

Annette Pilkington Lecture 25 : Integral Test R ∞ 2 R t 2 R 3t+5 2 I 1 3x+5 dx = limt→∞ 1 3x+5 dx = 7 3u du(u = 3x + 5) = ˛3t+5 2 ˛ 2 limt→∞ 3 ln |u| = limt→∞ 3 [ln |3t + 5| − ln |7|] = ∞. (integral ˛7 diverges). P∞ 2 I Therefore, the series n=1 3n+5 diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

∞ X 2 3n + 5 n=1

Annette Pilkington Lecture 25 : Integral Test P∞ 2 I Therefore, the series n=1 3n+5 diverges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

∞ X 2 3n + 5 n=1

2 I Consider the function f (x) = 3x+5 . I This function is positive and continuous on the interval [1, ∞). We see 0 −6 that it is decreasing by examining the derivative. f (x) = (3x+5)2 < 0. 2 I Since f (n) = 3n+5 , to determine convergence or divergence of the series P∞ 2 R ∞ n=1 3n+5 it suﬃces to check what happens to 1 f (x)dx. R ∞ 2 R t 2 R 3t+5 2 I 1 3x+5 dx = limt→∞ 1 3x+5 dx = 7 3u du(u = 3x + 5) = ˛3t+5 2 ˛ 2 limt→∞ 3 ln |u| = limt→∞ 3 [ln |3t + 5| − ln |7|] = ∞. (integral ˛7 diverges).

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

∞ X 2 3n + 5 n=1

Annette Pilkington Lecture 25 : Integral Test −x2 I Consider the function f (x) = xe .

I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

−n2 I Since f (n) = ne , to determine convergence or divergence of the series P∞ −n2 R ∞ n=1 ne it suﬃces to check what happens to 1 f (x)dx. 2 R ∞ −x2 R t −x2 1 R −t u I 1 xe dx = limt→∞ 1 xe dx = −2 limt→∞ −1 e du, (where 2 ˛−t 2 2 −1 u˛ −1 −t −1 1 u = −x ) = limt→∞ 2 e = limt→∞ 2 [e − e ] = 2e . (integral ˛−1 converges). P∞ −n2 I Therefore, the series n=1 ne converges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

Annette Pilkington Lecture 25 : Integral Test I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

−x2 I Consider the function f (x) = xe .

Annette Pilkington Lecture 25 : Integral Test −n2 I Since f (n) = ne , to determine convergence or divergence of the series P∞ −n2 R ∞ n=1 ne it suﬃces to check what happens to 1 f (x)dx. 2 R ∞ −x2 R t −x2 1 R −t u I 1 xe dx = limt→∞ 1 xe dx = −2 limt→∞ −1 e du, (where 2 ˛−t 2 2 −1 u˛ −1 −t −1 1 u = −x ) = limt→∞ 2 e = limt→∞ 2 [e − e ] = 2e . (integral ˛−1 converges). P∞ −n2 I Therefore, the series n=1 ne converges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

−x2 I Consider the function f (x) = xe .

I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

Annette Pilkington Lecture 25 : Integral Test 2 R ∞ −x2 R t −x2 1 R −t u I 1 xe dx = limt→∞ 1 xe dx = −2 limt→∞ −1 e du, (where 2 ˛−t 2 2 −1 u˛ −1 −t −1 1 u = −x ) = limt→∞ 2 e = limt→∞ 2 [e − e ] = 2e . (integral ˛−1 converges). P∞ −n2 I Therefore, the series n=1 ne converges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

−x2 I Consider the function f (x) = xe .

I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

−n2 I Since f (n) = ne , to determine convergence or divergence of the series P∞ −n2 R ∞ n=1 ne it suﬃces to check what happens to 1 f (x)dx.

Annette Pilkington Lecture 25 : Integral Test P∞ −n2 I Therefore, the series n=1 ne converges.

Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

−x2 I Consider the function f (x) = xe .

I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series Integral test, Example.

Example Use the integral test to determine if the following series converges:

∞ X −n2 ne n=1

−x2 I Consider the function f (x) = xe .

I This function is positive and continuous on the interval [1, ∞). We see that it is decreasing by examining the derivative. 2 2 2 f 0(x) = e−x − 2e−x x 2 = e−x (1 − 2x 2) < 0 if x ≥ 1.

Annette Pilkington Lecture 25 : Integral Test ∞ 1 P √ diverges since p = 1/3 < 1. I n=1 3 n P∞ 1 I n=1 n15 converges since p = 15 > 1. P∞ 1 I n=10 n15 also diverges since a ﬁnite number of terms have no eﬀect whether a series converges or diverges. ∞ 1 ∞ 1 P √ conv/diverges if and only if P √ conv/div. This diverges I n=100 5 n n=1 5 n since p = 1/5 < 1.

Integral Test Integral Test Example Integral Test Example p-series p-series

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Example Determine if the following series converge or diverge:

∞ ∞ ∞ ∞ X 1 X 1 X 1 X 1 √ , , , √ , 3 n n15 n15 5 n n=1 n=1 n=10 n=100

Annette Pilkington Lecture 25 : Integral Test P∞ 1 I n=1 n15 converges since p = 15 > 1. P∞ 1 I n=10 n15 also diverges since a ﬁnite number of terms have no eﬀect whether a series converges or diverges. ∞ 1 ∞ 1 P √ conv/diverges if and only if P √ conv/div. This diverges I n=100 5 n n=1 5 n since p = 1/5 < 1.

Integral Test Integral Test Example Integral Test Example p-series p-series

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Example Determine if the following series converge or diverge:

∞ ∞ ∞ ∞ X 1 X 1 X 1 X 1 √ , , , √ , 3 n n15 n15 5 n n=1 n=1 n=10 n=100

∞ 1 P √ diverges since p = 1/3 < 1. I n=1 3 n

Annette Pilkington Lecture 25 : Integral Test P∞ 1 I n=10 n15 also diverges since a ﬁnite number of terms have no eﬀect whether a series converges or diverges. ∞ 1 ∞ 1 P √ conv/diverges if and only if P √ conv/div. This diverges I n=100 5 n n=1 5 n since p = 1/5 < 1.

Integral Test Integral Test Example Integral Test Example p-series p-series

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Example Determine if the following series converge or diverge:

∞ ∞ ∞ ∞ X 1 X 1 X 1 X 1 √ , , , √ , 3 n n15 n15 5 n n=1 n=1 n=10 n=100

∞ 1 P √ diverges since p = 1/3 < 1. I n=1 3 n P∞ 1 I n=1 n15 converges since p = 15 > 1.

Annette Pilkington Lecture 25 : Integral Test ∞ 1 ∞ 1 P √ conv/diverges if and only if P √ conv/div. This diverges I n=100 5 n n=1 5 n since p = 1/5 < 1.

Integral Test Integral Test Example Integral Test Example p-series p-series

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Example Determine if the following series converge or diverge:

∞ ∞ ∞ ∞ X 1 X 1 X 1 X 1 √ , , , √ , 3 n n15 n15 5 n n=1 n=1 n=10 n=100

Annette Pilkington Lecture 25 : Integral Test Integral Test Integral Test Example Integral Test Example p-series p-series

R ∞ 1 We know that 1 xp dx converges if p > 1 and diverges if p ≤ 1.

∞ X 1 converges for p > 1, diverges for p ≤ 1. np n=1

Example Determine if the following series converge or diverge:

∞ ∞ ∞ ∞ X 1 X 1 X 1 X 1 √ , , , √ , 3 n n15 n15 5 n n=1 n=1 n=10 n=100

∞ 1 P √ diverges since p = 1/3 < 1. I n=1 3 n P∞ 1 I n=1 n15 converges since p = 15 > 1. P∞ 1 I n=10 n15 also diverges since a ﬁnite number of terms have no eﬀect whether a series converges or diverges. ∞ 1 ∞ 1 P √ conv/diverges if and only if P √ conv/div. This diverges I n=100 5 n n=1 5 n since p = 1/5 < 1.

Annette Pilkington Lecture 25 : Integral Test