Advanced Tests for Convergence

Sarah Fix

May 8, 2019

Abstract

The primary objective of this paper is to discuss advanced tests of convergence for

inﬁnite series. Commonly used tests for convergence that are taught to students in

early calculus classes, including the Comparison, Root, and Ratio Tests are not suﬃ-

cient in giving results for more complicated inﬁnite series. These frequently used tests

are discussed in the paper, along with examples of inﬁnite series that have interesting

properties, in order to eﬀectively examine the more advanced Kummer’s and Raabe’s

Tests. We demonstrate some applications of these more generalized tests through ex-

amples where simpler tests fail to yield results. While the main focus of this project is

on advanced tests for convergence, we also illustrate connections between the diﬀerent

tests.

Introduction

The study of inﬁnite series began when Archimedes used the ﬁrst summation of

an inﬁnite series to ﬁnd the area underneath an arc of a parabola. Long after Archimedes

wrote the ﬁrst summation, some of the most famous mathematicians of all time have worked

on inﬁnite series, including Gauss, Einstein, Euler, the Bernoulli brothers, and many more.

Throughout the centuries, the study of inﬁnite series has led to the proofs of some absolutely

astounding results. For example, in 1734, Leonhard Euler proved that the inﬁnite sum of

the inverse of squares is equal to

. This result was amazing not just because it is an

example of how adding up inﬁnitely many numbers can give a single, numerical result but

also because the result involves a seemingly unrelated constant, π. This key observation is

one of many that demonstrate the complexity and beauty that lies in the world of inﬁnite

series.

We begin by proving the commonly used Root and Ratio Tests. It is critical to

understand the origins and implications of these tests, especially the Ratio Test, as it is

the basis for one of our more complex tests. We also brieﬂy mention the “strength” of a

convergence test, and prove the relationship between the Root and Ratio Tests as well as

demonstrate this idea of “strength”.

An important piece of understanding inﬁnite series is recognizing the connection

between sequences and series. We address this connection through some examples of inﬁnite

series that have interesting properties. These examples include proofs that show convergence

or divergence in a variety of ways, including using the Cauchy Criterion for sequences and

“telescoping sums”. In addition, we use our previous knowledge on p-series and geometric

series to make comparisons in order to solve problems. As a resource for understanding

these properties of sequences and speciﬁc types of series, any background knowledge that we

need and all deﬁnitions excluding the advanced tests deﬁnitions can be found in most Real

Analysis textbooks such as Real Analysis: A First Course by Russell A. Gordon. Plus, any

additional information on the commonly used tests for convergence can be found in most

calculus books.

After we have reviewed some common aspects of inﬁnite series, we move on to

our more advanced tests of convergence, named after Ernst Eduard Kummer (1810-1893)

and Joseph Ludwig Raabe (1801-1859). Kummer was a German mathematician and Raabe

was a Swiss mathematician, both of whom did work on inﬁnite series at about the same

time in the mid-1800s. Raabe’s Test actually preceded Kummer’s Test; the latter is just

a generalized version of the former. We prove both of these tests as well apply them to

some examples and make a connection to one of our better-known tests. We then ﬁnish

our examination of these advanced tests with examples that employ both an advanced test

and other strategies previously mentioned in the ﬁrst section of the paper in order to tie

together all of the main themes of the project.

1 Review and Preliminaries

As previously mentioned, we start with the two convergence tests that are the

most common and easy to understand. While these tests and possibly even their proofs

may be familiar, it is important to truly understand these tests and where they come from

because they are the basis for some of the more advanced tests.

Theorem 1 (Ratio Test). Let

∞

n=1

be an inﬁnite series of nonzero terms and let

L = lim

n→∞



n+1



where it is assumed that the limit exists. Then

(i) if 0 ≤ L < 1, then the series

∞

n=1

converges;

(ii) if 1 < L ≤ ∞, then the series

∞

n=1

diverges;

(iii) if L = 1, then the test is inconclusive and the series may either converge or diverge.

Proof. We begin by considering the case where L < 1. Let r be a number such that

L < r < 1. Because L = lim

n→∞



n+1



and L < r, it follows that there exists an integer N

such that



n+1



< r ⇔ |a

n+1

| < |a

for all n ≥ N . Now, if we let n be equal to N , N + 1, N + 2, and so on, we ﬁnd that

N+1

| < |a

N+2

| < |a

N+1

|r < |a

N+3

| < |a

N+2

|r < |a

N+1

< |a

and, in general,

N+k

| < |a

for all positive integers k. The series

∞

k=1

is a geometric series with common ratio r

and because |r| < 1, the series

∞

k=1

must converge. Thus, by the Comparison Test,

the series

∞

k=N +1

| must also converge, it follows that

∞

n=1

| converges.

Now, consider the case where L > 1. Because L = lim

n→∞



n+1



and L > 1, there exists

some positive integer K such that



n+1



> 1

for all n ≥ K. This implies that

n+1

| > |a

| > 0

for all n ≥ K. This tells us that lim

n→∞

6= 0, and we conclude that the series diverges. 

Theorem 2 (Root Test). Let

∞

n=1

be an inﬁnite series of real numbers and let

ρ = lim

n→∞

when the limit exists. Then

(i) if 0 ≤ ρ < 1, then the series

∞

n=1

converges;

(ii) if 1 < ρ ≤ ∞, then the series

∞

n=1

diverges;

(iii) if ρ = 1, then the test is inconclusive and

∞

n=1

may either converge or diverge.

Proof. We ﬁrst consider the case when ρ < 1. Let r be a number such that ρ < r < 1.

Then there exists some positive integer N

such that

| < r for all n ≥ N

. Raising

both sides of the equation to the nth power gives us |a

| < r

for all n ≥ N

. Thus, because

∞

n=1

is a convergent geometric series, it follows that

∞

n=1

| converges by the Comparison

Test.

Now suppose that ρ > 1 and let r be a number such that 1 < r < ρ. Using similar

reasoning, we can see that there exists some positive integer N

such that for all n ≥ N

, we

have

| > r. Raising both sides to the nth power gives us |a

| > r

> 1 for all n ≥ N

Because

∞

n=1

is a divergent geometric series, we ﬁnd that the series

∞

n=1

| diverges by

the Comparison Test. 

As a preliminary, we consider the “strength” of convergence tests as part of our

discussion of the diﬀerent types of tests. When referring to the “strength” of a test, we

are referring to its ability to yield a result. For one convergence test to be “stronger” than

another, the stronger test giving a result implies that the weaker test also yields a result.

It must also be true that there exists a case where the stronger convergence test works but

the weaker test does not.

Theorem 3. The Root Test is “stronger” than the Ratio Test.

Proof. Without loss of generality, let a

> 0 for all positive integers n. Essentially, what

we want to prove is that if lim

n→∞



n+1



exists, then lim

n→∞

√

also exists and is equal. This

implies that if we are able to use the Ratio Test for a series, then the Root Test works as

well.

Let L = lim

n→∞



n+1



where 0 < L < ∞ and let  > 0. Then there exists some

integer N such that

L − <

n+1

< L + 

for all n ≥ N . It follows that

(L −)a

< a

n+1

< (L + )a

(L −)

< (L − )a

n+1

< a

n+2

< (L + )a

n+1

< (L − )

(L −)

< a

n+i

< (L + )

Then for n ≥ N we have

(L −)

< a

< (L + )

(L + )

and hence

L − <

√

< L + 

If the limit L exists, then because  > 0 was arbitrary we can conclude that lim

n→∞

√

= L.

Therefore, we have shown that the Root Test is at least as strong as the Ratio Test.

We claim that the Root Test is stronger than the Ratio Test and we prove this

by giving an example. To demonstrate this, we will show that there exist series such that

the Root Test indicates whether the series converges or diverges but the Ratio Test is

inconclusive.

Example. Consider the series

∞

n=1

where

(

if n is even,

n+1

if n is odd.

It follows that the ratio of consecutive terms will either be 1 or

, thus lim

n→∞

n+1

does not

exist. However, applying the Root Test gives us

lim

n→∞

√

= lim

n→∞

√

and lim

n→∞

√

= lim

n→∞

√

n+1

Therefore, the series converges by the Root Test while the Ratio Test is inconclusive and

thus we have shown that the Root Test is stronger than the Ratio Test. 

2 Sequences and Series

The purpose of the results and examples in this section are to illustrate some

interesting ﬁndings related to series and their sequences of partial sums. One important

result that will come up often is that if the sequence of partial sums of an inﬁnite series of

positive terms is bounded, then the series converges. (Gordon, Theorem 6.4). Therefore, if

a sequence of partial sums is not Cauchy, then the corresponding series does not converge.

Our ﬁrst set of results and examples considers a convergent inﬁnite series and its

sequence of partial sums. We demonstrate the use of sequences of partial sums to prove

convergence as well as our previous knowledge on the well known geometric and p-series.

Theorem 4. Let

∞

k=1

be a convergent series of positive terms and let t

∞

k=n

for each

positive integer n. The series

∞

k=1

diverges but the series

∞

k=1

√

converges.

Proof. We begin by noting that {t

} is a strictly decreasing sequence that converges to 0.

Therefore, given any positive integer m, there exists a positive integer n > 2m such that

m+1

. Then we have

k=m+1

m+1

k=m+1

−t

k+1

) >

m+1

n−1

k=m+1

−t

k+1

) =

m+1

− t

m+1

Hence, the sequence of partial sums for the series

∞

k=1

is not a Cauchy sequence. It follows

that

∞

k=1

diverges.

To show that the series

∞

k=1

√

converges, we ﬁrst note that

= t

− t

k+1



√

k+1



√

−

k+1



< 2

√



√

−

k+1



and thus

√

< 2



√

−

k+1



for all positive integers k. Using “telescoping sums”, we have

k=1

√

k=1



√

−

k+1



= 2



√

−

n+1



< 2

√

Thus, the sequence of partial sums for the series

∞

k=1

√

is bounded and therefore converges.

We conclude that the series

∞

k=1

√

converges. 

Example. To illustrate what we have just shown, we consider both of these series for

= 2

−k

and also for a

= (k

+ k)

−1

When a

, the series

∞

k=1

is a geometric series so we ﬁnd that t

. This

gives us

and

√

Thus the series

∞

k=1

∞

k=1

clearly diverges while the series

∞

k=1

∞

k=1

√



√



is a

geometric series with r =

√

< 1, hence the series converges.

When a

= (k

+ k)

−1

, the equation for t

is a “telescoping sum” so we have

∞

i=k

i(i + 1)

∞

i=k

(

−

i + 1

) =

It follows that

+ k

· k =

k + 1

for all k. Since

∞

k=1

k+1

diverges, the series

∞

k=1

diverges. Then considering the series

∞

k=1

√

, we have

√

+ k

√

k(k + 1)

for all k. We know that

∞

k=1

is a convergent p-series since

> 1. Hence, as expected,

the series

∞

k=1

√

converges.

Now we look at the case where the series

∞

n=1

is divergent. Note the similarity

to the results we have just shown.

Theorem 5. Let

∞

k=1

be a divergent series of positive terms and let {s

} be the sequence

of partial sums of this series. The series

∞

k=1

diverges but the series

∞

k=1

converges.

Proof. By the deﬁnition of s

, we see that a

n+1

= s

n+1

−s

, and because the series

∞

k=1

divergent the sequence {s

} is increasing and unbounded. Hence, given any positive integer

m there exists an integer n > 2m such that s

> 2s

. Thus we have

k=m+1

m+1

+ ···+

m+1

+ ···+ a

− s

= 1 −

> 1 −

Therefore, the sequence of partial sums is not Cauchy and hence the series

∞

k=1

diverges.

To show that the series

∞

k=1

converges, we use similar reasoning as in previous

problems. Noting that that the sum

k=2



k−1

−



telescopes to

−

we have

k=1

k−1

k=1

− s

k−1

k=1

k−1

−

for all k > 1. Hence, the sequence of partial sums is bounded so the series

∞

k=1

converges.



Example. Again to display the implications of this result, we consider both series for

convergence when a

= 1 and a

√

k + 1 −

√

When a

= 1, the equation for s

is simply s

= k. We then have

∞

k=1

∞

k=1

and

∞

k=1

∞

k=1

where

∞

k=1

is a divergent series and

∞

k=1

is a convergent series.

Because

k=1



√

k + 1 −

√



is a “telescoping sum”, the equation for s

i=1



√

i + 1 −

√



√

k + 1 − 1.

It follows that

√

k + 1 −

√

k + 1 − 1

√

k + 1 −

√

k + 1

(k + 1) +

√

k + 1

(k + 1) +

√

k + 1

√

k + 1

2(k + 1)

for all k ≥ 1. Therefore, since

∞

k=1

2k+2

diverges, the series

∞

k=1

√

k+1−

√

k+1−1

diverges by the

Comparison Test.

Next, we note that

√

k + 1 −

√

(

√

k + 1 − 1)

√

k + 1 −

√





√

k + 1 +

√

k + 1 +

√





√

k + 1 +

√







√



We know that the inﬁnite series

∞

k=1

converges because it is a p-series where p =

. Thus,

by the Comparison Test, the series

∞

k=1

√

k+1−

√

(

√

k+1−1)

must also converge.

The last result in this section also demonstrates the relationship between sequences

and series but instead of starting with a series and using its sequence of partial sums in our

proof, we begin with a sequence and make the connection to the corresponding series of the

terms of the sequence.

Theorem 6. Suppose the sequence {

√

k · a

} converges to a positive number. Then the

series

∞

k=1

converges while the series

∞

k=1

diverges.

Proof. Because {

√

k · a

} is a convergent sequence, this implies that it is also bounded.

Thus, there exists some number M such that |

√

k · a

| ≤ M for all k ≥ 1. Dividing both

sides of the inequality by k

yields the result



≤

for all k ≥ 1. Since

∞

k=1

is a convergent p-series, it follows that

∞

k=1

converges by the

Comparison Test.

Since the sequence {

√

k · a

} converges to a positive number, we know that it is

eventually bounded below by some positive number α. This means that there exists some

integer K such that

0 < α ≤

√

k · a

for all k ≥ K. Dividing both sides by

√

k yields the result

√

≤ a

for all k ≥ K. Then, we know that

∞

k=1

√

diverges and thus the series

∞

k=K

diverges by

the Comparison Test. Hence, the series

∞

k=1

also diverges. 

3 More Advanced Tests

Now that we have covered some interesting properties of certain series, we examine

some more advanced tests for convergence. While we can see that the techniques we have

used thus far are useful and can yield intriguing results, the tests and strategies we have

just displayed do not always determine convergence and divergence for a series. In fact, it

happens often that the tests we mentioned give us no information. To illustrate this, we

consider a generic p-series and apply the Ratio and Root Tests:

Let a

and note that

lim

n→∞

n+1

= lim

n→∞

n + 1

= 1

and

lim

n→∞

√

lim

n→∞

= lim

n→∞

−p

)

= lim

n→∞



√



−p

= 1.

We know that a p-series diverges when p ≤ 1 and converges when p > 1. Therefore, we

cannot determine whether the series converges or diverges, because both the Ratio and Root

Test gives us 1 no matter what p is. In general, these tests fail when they give a result of 1

because these tests compare the series to a geometric series with r equal to the result of the

test. When r = 1, the terms of a geometric series are constant. Even though the geometric

series would diverge in this, case as a comparison we get no information on whether the

terms of the series are going oﬀ to inﬁnity or converging.

As we can see, the Root and Ratio Test are often not helpful in deciphering the

convergence of an inﬁnite series. In the case of the p-series we can often use the Comparison

Test, but when we want to examine more complicated series using the Comparison Test can

get tricky. This is why we need more advanced tests like the ones that follow.

Theorem 7 (Kummer’s Test). Let

∞

n=1

be a series of positive terms and let {B

} be a

be a sequence of positive constants. Let

L = lim

n→∞

n+1

− B

n+1

where it is assumed that the limit exists.

(i) If L > 0, then the series

∞

n=1

converges.

(ii) If L < 0 and

∞

n=1

diverges, then the series

∞

n=1

diverges.

Proof. We begin by considering the case where L > 0. Choose a number r such that

0 < r < L. Then there must exist some integer N > 0 such that

n+1

− B

n+1

> r ⇔ B

− B

n+1

> r a

n+1

for all n ≥ N . Given any positive integer m, we have

− B

N+1

> r a

N+1

− B

N+2

> r a

N+2

− B

N+3

> r a

N+3

N+m−1

− B

N+m

> r a

N+m

And then by adding these inequalities together, we get cancellations of all of the terms on

the left side of the inequality except for the ﬁrst and last. Thus we have

− B

N+m

> r (a

N+1

+ ··· + a

N+m

) = r (S

N+m

− S

where S

is the partial sum for

N+m

k=N +1

. It follows that

N+m

< r S

+ B

− B

N+m

< r S

+ B

Let C be the constant

(r S

)

. The above inequality tells us that S

< C for all n ≥ N.

Thus, the sequence of partial sums {S

} of the series

∞

n=1

is bounded and therefore the

series converges.

Now consider the case where L < 0. This means that there exists an integer N > 0

such that B

n+1

− B

n+1

≤ 0 for all n ≥ N . Rearranging this inequality gives us

≤ B

n+1

for all n ≥ N which then implies that

≤ B

for all n ≥ N . Letting B

be a constant C, we have

≥

for n ≥ N and, because

∞

n=1

diverges, the series

∞

n=1

diverges by the Comparison Test.



Remark. Using Kummer’s Test, if we let the sequence {B

} = {1}, then we get

L = lim

n→∞

n+1

− 1

If lim

n→∞



n+1



> 1, then it follows that L is positive and thus the series converges by

Kummer’s Test. This also tells us that the ratio lim

n→∞



n+1



< 1, and thus the series

converges by the Ratio Test. Similarly, if lim

n→∞



n+1



< 1 then L is negative and the series

diverges by Kummer’s Test. From this we can see that the ratio lim

n→∞



n+1



> 1 so we

conclude that the series also diverges by the Ratio Test. Therefore, we can conclude that

the Ratio Test is simply Kummer’s Test associated with the sequence {B

} = {1}

Kummer’s Test is inconclusive when L = 0, but when this happens it is possible to

reﬁne the sequence {B

} to give a result, which is why the case when L = 0 is not included

in the deﬁnition of the test. But because there is no speciﬁc strategy for choosing {B

this can become a tedious and time-consuming task in order to ﬁnd a sequence that yields

a result when used in Kummer’s Test. This is where Raabe’s Test comes in. Raabe’s Test

is simply a speciﬁc case of Kummer’s Test where we let B

= n, therefore we do not have

to go through the process of trying to ﬁnd a suitable B

Theorem 8 (Raabe’s Test). Let

∞

n=1

be a series with positive terms and assume that

ρ = lim

n→∞

n+1

− 1

exists. Then

(i) if ρ > 1, then the series

∞

n=1

converges.

(ii) if ρ < 1, then the series

∞

n=1

diverges.

(iii) if ρ = 1, then

∞

n=1

may either converges or diverge and the test is inconclusive.

Proof. Items (i) and (ii) are consequences of Kummer’s Test. To demonstrate this, let

= n and compute

L = lim

n→∞

n+1

− (n + 1)

= lim

n→∞



n+1

− 1



− 1

= ρ − 1

If ρ > 1, then L > 0 and the series

∞

n=1

converges by Kummer’s Test. Similarly, if ρ < 1,

then L < 0 and note that

∞

n=1

diverges, thus the series

∞

n=1

diverges by Kummer’s Test.

If the result of Raabe’s Test is ρ = 1, then the test is inconclusive. 

Remark. Applying Raabe’s Test to the series

∞

n=1

gives us

ρ = lim

n→∞



(n + 1)

− 1



= lim

n→∞





n + 1



− n



= lim

n→∞



1 +



− 1

= p

Thus, if ρ > 1, then the series converges by Raabe’s Test. This implies that p > 1 which

indicates that the series is a convergent p-series. On the other hand, if ρ < 1 then we know

the series diverges by Raabe’s Test, and this implies that p < 1 and therefore the series is

a divergent p-series. This demonstrates that Raabe’s Test tests a series for convergence by

comparing the series to the p-series where p = ρ.

Corollary 1 (Of Raabe’s Test). In general, when ρ > 0, the sequence {a

} converges to 0.

Proof. When ρ > 0, there must exist some integer M such that



n+1

− 1



> 0

for all n ≥ M . It follows that a

> a

n+1

for all n ≥ M . This tells us that {a

} is eventually

decreasing, therefore {a

} converges to some number L.

To show that L = 0, we use proof by contradiction and assume that L > 0. Similar to our

above reasoning, because ρ > 0 there exists some integer N and a number  > 0 such that



n+1

− 1



> 

for all n ≥ N . From this, we see that

n+1

− 1 >



=⇒ a

− a

n+1



n+1

 L

Now, “telescoping” this diﬀerence gives us

− a

= (a

− a

n+1

) + (a

n+1

− a

n+2

) + ··· + (a

m−1

− a

)

>  L



n + 1

+ ··· +

m − 1



whenever m > n ≥ N. Since the sequence of partial sums for the harmonic series diverges

and we know that {a

} converges, we have reached a contradiction. Therefore, the sequence

} converges to 0. 

As previously mentioned, Raabe’s Test chronilogically preceded Kummer’s Test,

so instead of considering Raabe’s Test to be a reﬁnement of Kummer’s Test, we could also

consider Kummer’s Test to be a generalized version of Raabe’s Test. The examples that

follow require Raabe’s Test to solve as the Root Test would be diﬃcult to calculate because

there are no powers of n, the Ratio Test fails, and the unique usage of the double factorial

makes the Comparison Test diﬃcult to use.

4 Applications of More Advanced Tests

The double factorial of an integer n denoted by n!! is the product of all the integers from

1 up to and including n that have the same parity as n. For example, 9!! = 9·7·5·3·1 = 945.

Example. Test the series

∞

n=1

(2n−1)!!

(2n)!!

2n+1

for convergence.

Let a

represent the terms of the series. Since

lim

n→∞

n+1

= lim

n→∞

(2n + 1)!!

(2n + 2)!!(2n + 3)

(2n)!!(2n + 1)

(2n − 1)!!

= lim

n→∞



2n + 1

2n + 2



2n + 1

2n + 3



= 1

the Ratio Test fails and we turn to Raabe’s Test. We ﬁnd that

lim

n→∞

n+1

− 1

= lim

n→∞

(2n − 1)!!(2n + 2)!!(2n + 3)

(2n + 1)!!(2n)!!(2n + 1)

− 1

= lim

n→∞

(2n + 2)(2n + 3)

(2n + 1)(2n + 1)

− 1

= lim

n→∞

+ 10n + 6

+ 4n + 1

− 1

= lim

n→∞

6n + 5

+ 4n + 1

= lim

n→∞

+ 5n

+ 4n + 1

which shows that the series converges by Raabe’s Test.

Example. Test the series

∞

n=1

(2n−1)!!

(2n)!!

4n+3

2n+2

for convergence.

Let a

represent the terms of the series. Using the Ratio Test gives us

lim

n→∞

n+1

= lim

n→∞

(2n + 1)!!(4n + 7)

(2n + 2)(2n + 4)

(2n)!!(2n + 2)

(2n − 1)!!(4n + 3)

= lim

n→∞



2n + 1

2n + 2



4n + 7

4n + 3



2n + 2

2n + 4



= 1

so again we turn to Raabe’s Test. We then have

lim

n→∞

n+1

− 1

= lim

n→∞

(2n − 1)!!(4n + 3)

(2n)!!(2n + 2)

(2n + 2)!!(2n + 4)

(2n + 1)!!(4n + 7)

− 1

= lim

n→∞



2n + 2

2n + 1



4n + 3

4n + 7)



2n + 4

2n + 2



− 1

= lim

n→∞

+ 22n + 12

+ 18n + 7

− 1

= lim

n→∞

+ 5n

+ 18n + 7

Thus the series diverges by Raabe’s Test.

While it does not have a name, the following version of Kummer’s Test, where

we let B

= n ln n, is similar to Raabe’s and is helpful when Raabe’s Test fails as in the

example that follows.

Theorem 9. Let

∞

n=1

be a series with positive terms and assume that

δ = lim

n→∞

ln n



n+1

− 1



− 1

exists. Then,

(i) if δ > 1, then the series

∞

n=1

converges.

(ii) if δ < 1, then the series

∞

n=1

diverges.

(iii) if δ = 1, then

∞

n=1

may either converge or diverge and the test is inconclusive.

Proof. To begin, we refer to Kummer’s Test and let B

= n ln n. Note that the series

∞

n=1

n ln n

diverges. Then we have

L = lim

n→∞

n ln n

n+1

− (n + 1) ln (n + 1)

= lim

n→∞

n ln n

n+1

− (n + 1) ln n − (n + 1) ln



1 +



= lim

n→∞

ln n

n+1

− (n + 1)

− (n + 1) ln



1 +



= lim

n→∞

ln n





n+1

− 1



− 1



− n ln



1 +



− ln



1 +



= lim

n→∞

ln n





n+1

− 1



− 1



− ln



1 +



− ln



1 +



Since lim

n→∞



1 +



= e, it follows that lim

n→∞



1 +



= 1 and lim

n→∞



1 +



= 0.

Thus we can see that L = δ − 1. This tells us that if δ > 1 then L > 0 and the series

converges by Kummer’s Test. Similarly, if δ < 1, then L < 0 and the series diverges. If

δ = 1 this means that L = 0 and the test is inconclusive, therefore a reﬁnement of Kummer’s

Test is necessary to get a result.

Example. Test the series

∞

n=1

(2n−1)!!

(2n)!!

for convergence.

Let a

represent the terms of the series. The Ratio Test gives us

lim

n→∞

n+1

= lim

n→∞



(2n + 1)!!(2n)!!

(2n − 1)!!(2n + 2)!!



= lim

n→∞



(2n + 1)

(2n + 2)



= 1

Once again, we turn to Raabe’s Test and we ﬁnd that

lim

n→∞

n+1

− 1

= lim

n→∞



2n + 2

2n + 1



− 1

Letting x =

2n+1

, it follows that n =

1−x

and thus because x approaches 0 as n approaches

inﬁnity we have

lim

n→∞



2n + 2

2n + 1



− 1

= lim

x→0

1 − x



(1 + x)

− 1



= lim

x→0

1 − x

(1 + x)

− 1

Using the product rule for limits, because both limits exist we can divide up the limit of

the product into the product of the limits of each fraction. Then we have

lim

x→0

1 − x

· lim

x→0

(1 + x)

− 1

· p =

Therefore we can see that if p > 2, the series converges and the series diverges when p < 2.

However, when p = 2 both the Ratio Test and Raabe’s Test fail. We use the previously

stated reﬁnement of Raabe’s Test which gives us

lim

n→∞

ln n



n+1

− 1



− 1

= lim

n→∞

ln n



2n + 2

2n + 1



− 1



− 1

= lim

n→∞

ln n



+ 8n + 4

+ 4n + 1

− 1



− 1

= lim

n→∞

ln n



4n + 3

+ 4n + 1



− 1

= lim

n→∞

ln n

+ 3n

+ 4n + 1

− 1

= lim

n→∞

ln n

−n − 1

+ 4n + 1

and because lim

n→∞

ln n



−n−1

+4n+1



= 0 < 1, the series diverges when p = 2.

Our ﬁnal examples will require us to use both Raabe’s Test and what we know

about some other common types of series; Power Series and Alternating Series.

Example. Find the interval of convergence for the power series

∞

k=1

1·4·7···· ·(3k−2)

To begin, we use the Ratio Test to determine the radius of convergence for the

series. The Ratio Test yields the result

lim

k→∞



1 · 4 · 7 · ··· · (3k + 1)

(k + 1)!

k+1

1 · 4 · 7 · ··· · (3k − 2) · x



= lim

k→∞



(3k + 1)

(k + 1)

· x



= 3|x|.

This tells us that the series converges when |3x| < 1. This means that the series converges

when the inequality −

< x <

is true. Now we must test the endpoints of this interval

for convergence at these points.

First, we consider when x =

. This results in

∞

k=1

1 · 4 · 7 · ··· · (3k − 2)





Letting a

1·4·7···· ·(3k−2)

k!·3

and applying Raabe’s Test we then have

lim

k→∞



k+1

− 1



= lim

k→∞



(k + 1)

(3k + 1)

· (3) − 1



= lim

k→∞



3k + 3

3k + 1

− 1



= lim

k→∞



3k + 1



= lim

k→∞



3k + 1



Therefore the series diverges when x =

by Raabe’s Test.

Now we consider the case where x = −

. The series

∞

k=1

(−1)

has negative terms,

so we cannot use Raabe’s Test. We now turn to the Alternating Series Test where a

the same as above. The Alternating Series Test tells us that if {a

} is decreasing and the

terms go to 0 then the series converges. In our above results, ρ =

> 0 so by Corollary 1,

the sequence {a

} goes to 0. Additionally, a

is always positive and so the sequence must

be decreasing. Therefore, the series

∞

k=1

1··4·7·...·(3k−2)



−



converges by the Alternating

Series Test. Hence, the interval of convergence for the power series

∞

k=1

1·4·7·...·(3k−2)



−



So far, our discussion on inﬁnite series has only touched on tests that give us

information on whether the series converges or not. If we ﬁnd that a series converges, these

tests do not give us a good indication as to what the sum of the series is. However, in our

last example we will illustrate how we can use the information that Raabe’s Test gives us

to ﬁnd the sum of a series.

Example. Find the sum of the series

∞

k=1

2·4·6···· ·(2k)

5·7·9···· ·(2k+5)

We begin by letting

2 · 4 · 6 · ··· · (2k)

5 · 7 · 9 · ··· · (2k + 5)

and b

2 · 4 · 6 · ··· · (2k)

3 · 5 · 7 · ··· · (2k + 3)

It follows that



2k + 2



k+1

and b



2k + 5

2k + 2



k+1

From here, we apply Raabe’s Test to the series

∞

k=1

which give us

lim

k→∞



k+1

− 1



= lim

k→∞



(2k + 5)

(2k + 2)

− 1



= lim

k→∞



2k + 2



Thus the series

∞

k=1

converges by Raabe’s Test.

Then, we observe that

− a



2k + 5

2k + 2



k+1

−



2k + 2



k+1



2k + 2



k+1

= b

k+1

We can use this result to conclude that

∞

k=1

= lim

n→∞

k=1

= lim

n→∞

k=1

− b

k+1

) = lim

n→∞

− b

n+1

) = b

Thus, the sum of the series

∞

k=1

2·4·6···· ·(2k)

5·7·9···· ·(2k+5)

Conclusion

The scope of this project merely skims the surface of what has been done in the

ﬁeld of inﬁnite series and tests of convergence. Beyond Kummer’s Test and Raabe’s Test,

there are many more tests including Bertrand’s, Gauss’s, and Dirichlet’s which are all useful

in their own way. Tests for convergence are important to the study of inﬁnite series because

it is often the case where knowing the property of the series is imperative to a proof or the

solution to a problem as shown in our last example.

In mathematics, some of the most useful applications of inﬁnite series are in inte-

gration and diﬀerential equations. We also use speciﬁc inﬁnite series such as Taylor Series

to show that calculations involving functions such as e

and sin x can be computed us-

ing addition, subtraction, multiplication, and division. Aside from mathematics, inﬁnite

series are also useful in a wide range of ﬁelds including physics, engineering, chemistry,

computer science, and ﬁnance. In physics they are used to solve the pendulum diﬀerential

equation while in ﬁnance they are used to calculate ﬁscal multipliers. Inﬁnite series also

have a wide range of uses in computer science as they are often used in generating functions.

Without inﬁnite series and the tests we use to determine convergence, we would not be able

to advance and solve important problems in many areas of study.

In this project we were able to show some very interesting results involving inﬁnite

series and their partial sums as well as draw some connections between series and sequences.

We introduced two advanced tests of convergence and demonstrated their usefulness as well

as their applications and relationships to one another. Further inquiry into this topic would

include the examination of other advanced tests such as the ones just mentioned and a

more rigorous investigation into the results and implications of the tests we discussed. In

conclusion, the study of inﬁnite series and tests of convergence is not only extensive, but it

is intriguing and beautiful.

Acknowledgements

The author would like to thank Professor Russell A. Gordon not only for his time

and eﬀort that he put into this paper but also for his continuous work with the author that

made this project possible. She would also like to thank Sarah Vesneske for proof-reading

and editing this paper time and time again.

References

[1] John M. H. (John Meigs Hubbell) Olmsted. Real Variables, an Introduction to the

Theory of Functions. Appleton-Century-Crofts, New York, 1959.

[2] Russell A. Gordon. Real Analysis: A First Course. 2nd ed. Boston: Addison-Wesley,

2002.

[3] Walter Rudin. Principles of Mathematical Analysis. Third ed. International Series in

Pure and Applied Mathematics. New York: McGraw-Hill, 1976.

[4] Brian S. Thomson, Judith B. Bruckner, and Andrew M. Bruckner. Elementary Real

Analysis. Second ed. Prentice Hall (Pearson), 2001.

[5] Robert Gardner Bartle. The Elements of Real Analysis. 2d ed. New York: Wiley, 1976.

[6] I. I. Hirschman. Inﬁnite Series. Athena Series. New York: Holt, Rinehart and Winston,

1962.

[7] James Stewart. Calculus : Early Transcendentals. 7th ed. Belmont, Cal.: Brooks/Cole,

Cengage Learning, 2012.