Convergence Tests

We are now interested in developing tests to tell whether or nota series converges, without having to guess at its sum. The first such test is entirely a negative result.

Theorem. (the divergence test).

(a) If , then the series diverges.

(b) If , then the series may either converge

or diverge.

lim 0ukk

≠→∞

uk∑

lim 0ukk

=→∞

uk∑

The divergence test is strictly a negative test. It can be used to show that a series diverges, but it cannot be used to show that a series converges.

Proof. (a) It is clear by definition that uk = sk − sk-1. If the series

converges to s, then we have lim lim .1

s s sk kk k

= =−→∞ →∞

Thus lim lim ( ) lim lim 0.1 1

u s s s s s sk k k k kk k k k

= − = − = − =− −→∞ →∞ →∞ →∞

(b) Follows by showing both a convergent series and a divergent series for which lim 0.u

kk=

→∞The two series are

1 1 112 4 2k

+ + + + +L L and 1 1 112 3 k

+ + + + +L L

Problem. (a) Show that the series

diverges.

2 2 4 6 22 3 4 5 21

k kk kk

∞= + + + + +∑ + +=

L L

2 2lim lim 2 022 1

kkk k

k

= = ≠+→∞ →∞ +

Solution. The nth term sequence has the limit

Therefore the series diverges.

Problem. (a) Show that the series

diverges.

1 3 5 9 11 12 4 82 21

k kk

∞ + = + + + + + +∑

=L L

1lim 1 1 02kk

+ = ≠→∞

Solution. The nth term sequence has the limit

Therefore the series diverges.

Theorem(a) If Σuk and Σvk are convergent series, then Σ(uk + vk) and Σ(uk − vk) are also convergent, and we have

(b) If c is a nonzero constant, then the series Σuk and Σcuk both converge or both diverge. In the case of convergence we have

( ) 1 1 1

u v u vk k k kk k k

∞ ∞ ∞+ = +∑ ∑ ∑

= = =

( ) 1 1 1

u v u vk k k kk k k

∞ ∞ ∞− = −∑ ∑ ∑

= = =

1 1cu c uk k

k k

∞ ∞=∑ ∑

= =

Example. Find the sum of the series5 7 13 41k k

k

∞−∑ −

=

The series breaks down into the difference of two series,

5 7 and .13 41 1k k

k k

∞ ∞∑ ∑ −= =

The first series 5 5 5 5 53 9 273 31

k kk

∞= + + + +∑

=L L

is geometric with a = 5/3 and r = 1/3. It therefore converges to

5 53 .1 21 3

=−

The second series 7 7 771 4 1641k

k

∞= + + +∑ −

=L is also

geometric with a = 7 and r = 1/4. It therefore converges to

7 7 28.1 3 31 4 4

= =−

Thus by (a) the original series converges to

5/2 − 28/3 = (15 − 56)/6 = −41/6.

Solution.7 1

7 .11k kkk

∞ ∞= ∑ ∑

== We showed before that the series

1

1kk

∞ ∑

= diverges. This series is a constant multiple of a

divergent series and so diverges by part (b).

Example. Determine whether the following series converges or diverges.

7 7 7 7 72 3 41kk

∞= + + + +∑

=L

TheoremConvergence or divergence is not affected by deleting a finite number of terms from a series; in particular, for any positive K, the series

and

both converge or both diverge.

1 2 3

1u u u uk

k

∞= + + +∑

=L

1 2

u u u uk K K Kk K

∞= + + +∑ + +

=L

Warning. The above theorem says that convergence is unaffected by removal of a finite number of terms. However, the sum usually is changed.

Example. Determine whether the following series converge or diverge.

(a) (b)1 1 1 1 8 9 108kk

∞= + + +∑

=L

Solution. (a) is just the divergent series

1

1kk

∞ ∑

= With the first 7 terms missing. By the theorem above

1 1 1 1 8 9 108kk

∞= + + +∑

=L

above, it must also diverge.

1 1 14 3 5 72 4 8

+ − + + + + +L

Example. Determine whether the following series converge or diverge.

(a) (b)1 1 1 1 8 9 108kk

∞= + + +∑

=L

1 1 12 4 8

+ + +LSolution. (b) The series is a geometric series with

a = 1/2 and r = 1/2. It therefore converges. Thus by the previous theorem the sequence

1 1 14 3 5 72 4 8

+ − + + + + +L

1 1 14 3 5 72 4 8

+ − + + + + +L

must also converge. Of course the sums are different. The geometric series formed when the first 4 terms are dropped has sum (1/2)/(1 − 1/2) = 1, and therefore the original series converges to 9 + 1 = 10.

The integral test

Let us compare the series with the improper integral 12

1kk

∞∑=

1.2

1dx

x

∞∫

Clearly, if the series converges, the sum of the series is greater than the integral, while if the integral diverges, so does the series.

1

1/41/9

On the other hand, the diagram below shows that if the integral

converges, so does the series and therefore so does the

original series. On the other hand, if the series diverges, then so must the integral.

12

2kk

∞∑=

1/41/9

1/16

Theorem (The Integral Test)Suppose that a series Σuk = Σf(k) has positive terms, and that f(x) is the formula resulting from replacing k by x in the expression for the terms of the series.

If the function with formula f(x) is decreasing and continuous on

the interval [a, ∞), then the series and the integral

both converge or both diverge.

This leads to the following theorem.

1uk

k

∞∑=

( )f x dxa

∞∫

Note that the value of a does not matter.

Example. Use the integral test to determine whether the following series converge or diverge.

1(a) 1kk

∞∑=

1(b) 21kk

∞∑=

1(c) , a positive number.1

ppkk

∞∑=

Solution (a). lim lim ln( ) .1 1

rdx dxr

x xr r

∞= = =∞∫ ∫

→∞ →∞

Since the improper integral converges, so does the series.

Example. Use the integral test to determine whether the following series converge or diverge.

1(a) 1kk

∞∑=

1(b) 21kk

∞∑=

Solution (b). 1 1lim lim lim 1 1.2 2 11 1

rrdx dxx rr r rx x

∞ = = − = − =∫ ∫ →∞ →∞ →∞

Since the improper integral converges, so does the series.

Note. The series does not converge to 1. The theorem does not say that the series and the improper integral have the same value when they converge.

1(c) , a positive number.1

ppkk

∞∑=

Example. Use the integral test to determine whether the following series converge or diverge.

1(a) 1kk

∞∑=

1(b) 21kk

∞∑=

Solution (c). We already know that this diverges if p = 1. If pis not 1, then we have

1 11lim lim lim 11 111 1

1 if 1 1

if 1

p rrdx xp px dx rp p pr r rx

pp

p

− ∞ − − = = = −∫ ∫ − − →∞ →∞ →∞

> −= ∞ <

1(c) , a positive number.1

ppkk

∞∑=

Theorem (The p - series test)

converges if p > 1 and diverges if 0 < p ≤ 1.

Thus we have the following result of using the integral test

on series .It is called the p - series test.1

1pkk

∞∑=

1 1 1 112 31

p p p pk kk

∞= + + + + +∑

=L L

Problem. Use any method to determine if the following series converges or diverges.

351 kk

∞∑=

Solution. The first thing we usually do is check the divergence test. Clearly the terms in this series do converge to 0, so the divergence test gives us no information.

We can write the series as 3 13

5 5 11 k kkk

∞ ∞= ∑ ∑

== Since the series in the brackets diverges, so does the series

351 kk

∞∑=

Problem. Use any method to determine if the following series converges or diverges.

1

1kk

∞∑=

Solution 2. The series is a p - series, and so by the p -

series test it diverges.

351 kk

∞∑=

351 kk

∞∑=

is a constant multiple of this p - series, so also diverges.

Problem. Use any method to determine if the following series converges or diverges. 2 1

2 31

k

kk

∞ +∑

+=

Solution. Again check the divergence test. In this case we have

Thus by the divergence test, the series diverges.

112 21lim lim 1 0

2 33 1 2

k kk kk

k

++ = = ≠

→∞ →∞+ +

2 12 31

k

kk

∞ +∑

+=

Problem. Use any method to determine if the following series converges or diverges.

ln

3

kkk

∞∑=

Solution. Here the function is integrable and

satisfies the conditions of the integral test. Thus the series behaves exactly as does the improper integral

ln( ) xf xx

=

ln lnlim

3 3

rx xdx dx

x xr

∞=∫ ∫

→∞

We pause to evaluateln

.x

dxx∫ Let u = ln x, du = dx/x.

Therefore the series diverges.

[ ]2 2ln ln 1 ln( ) ln(3)2lim lim ln( ) lim

2 2 233 3

rrx x rdx dx x

x xr r r

∞ = = = − =∞ ∫ ∫ →∞ →∞ →∞

Then 2ln 21 ln( )2 2

x udx udu xx

= = = ∫ ∫

Continuing with the previous equation , we have

Problem. Use any method to determine if the following series converges or diverges. 2

1

kkek

∞ −∑=

Solution. Here the function is integrable and

satisfies the conditions of the integral test. Thus the series behaves exactly as does the improper integral

2( ) xf x xe−=

2 2lim

1 1

rx xxe dx xe dxr

∞ − −=∫ ∫→∞

We pause to evaluate2

.xxe dx−∫ Let u = x2, du = 2xdx.

Therefore the series converges (but not to 1/2e).

2 2 21 11lim lim2 21 1

rx x rxe dx xe dx e eer r

∞ − − − −= =− − = ∫ ∫→∞ →∞

Then 2 21 1 1

2 2 2x u u xxe dx e du e e− − − −= =− =−∫ ∫

Continuing with the previous equation , we have

Problem. Use any method to determine if the following series converges or diverges.

1 1 1 1 11 3 4 5 12 k kk

∞= + + + + +∑ + +=

L L

Solution 1. This series is a p - series with p = 1/2, with 2 terms dropped off. Thus it diverges.

Solution 2. We can use the integral test directly.

lim lim 2 1 lim 2 1 2 31 1 22 2

rrdx dx x rx xr r r

∞ = = + = + − =∞∫ ∫ + +→∞ →∞ →∞

We see again that the series must diverge.