H.A.E.F. Psychico College Stylianos RousoglouCandidate number: 000901-0086

EXTENDED ESSAY in MATHEMATICS

A Study of More Advanced Series Convergence Tests and Treating Tail Behavior

Total word count: 2979

MAY 2014

000901-0086

Abstract

This work is a study of advanced series convergence tests that can deal with

complicated series and treat tail behavior. I first define, prove and examine in detail the

series convergence tests touched upon in the IB option Calculus Curriculum, namely the

Comparison test (limit and direct form), the Integral test and the Ratio test. Since proofs

of these tests are not covered by the syllabus, I focus on them, while also giving

examples of how the tests are applied to determine the convergence or divergence of

series. Furthermore, I explicitly show cases where the above tests fail to provide

concrete answers, therefore creating the need for more advanced series convergence

tests. I then present a generalization of the Ratio Test, namely the Root test, and give a

concrete example of why the former may fail while the latter may provide a definite

answer (but not vice-versa). By proving and thoroughly studying these tests, I create the

mathematical background necessary for me to proceed in investigating more complex

and advanced convergence tests.

The second part of the work focuses on three advanced series convergence tests,

namely Kummer’s test, Raabe’s rest, and Gauss’s test. The former is employed to aid in

the proofs of Raabe’s and Gauss’s tests, while the later is only deduced after using both

Kummer’s and Raabe’s tests. I then provide concrete examples, through which the

importance of the tests is evident, since we are now capable of dealing with series of

more than one variable, as well as summing piece-wise sequences. By treating a much

wider spectrum of series, these three widely known series convergence tests are an

important addition to the tools we already posses for determining the convergence or

divergence of the summation of a sequence.

Word count: 291

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Table of Contents

Introduction.............................................................................................................................. 1

Two Fundamental Proofs....................................................................................................... 2

Elementary Convergence Tests............................................................................................. 3Proposition 1 (Comparison Test)....................................................................................................... 3

Proof.................................................................................................................................................................... 3Application 1....................................................................................................................................................3

Proposition 2 (Limit Comparison Test)............................................................................................ 4Proof.................................................................................................................................................................... 4Application 2....................................................................................................................................................5

Proposition 3 (Integral Test)...............................................................................................................5Proof.................................................................................................................................................................... 5Application 3....................................................................................................................................................6Application 4....................................................................................................................................................6

Proposition 4 (Ratio Test)...................................................................................................................7Proof.................................................................................................................................................................... 7

Proposition 5 (Root Test).................................................................................................................... 9Proof.................................................................................................................................................................... 9Application 5.................................................................................................................................................10

Advanced Convergence Tests............................................................................................. 11Proposition 6 (Kummer’s Test)....................................................................................................... 11

Proof of (i):.....................................................................................................................................................11Proof of (ii):....................................................................................................................................................12

Proposition 7 (Raabe’s Test)............................................................................................................13Proof..................................................................................................................................................................13Alternate form of Raabe's test (Limit form):...................................................................................14Application 6.................................................................................................................................................14

Proposition 8 (Gauss’s Test)............................................................................................................15Proof..................................................................................................................................................................15Application 7.................................................................................................................................................17Application 8.................................................................................................................................................18

Conclusion............................................................................................................................. 19

Bibliography......................................................................................................................... 20

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Introduction

The IB Mathematics Higher Level option Calculus curriculum covers some

fundamental tests for convergence of infinite series, for example the Comparison test

(limit and direct form), the Integral test and the Ratio test. Nevertheless, all the above

tests cover a relatively limited spectrum of series that usually depend on one variable.

Even for the case of simple series that these tests treat, there exist examples for which

they are inconclusive. Furthermore, if the sequence being summed depends on more than

one parameter, these tests are practically ineffective. Therefore, a study of more

advanced convergence tests is needed.

As the coverage of the basic convergence tests in the Syllabus is superficial, I

start by proving in detail the major tests contained in the Option Calculus, to build the

necessary mathematical infrastructure on which I will base my investigation of more

advanced convergence tests. In the second part of the work, I state, prove and provide

concrete examples for the use of such series convergence techniques, namely Kummer’s,

Raabe’s, and Gauss’s series convergence tests. These tests are tools that can be used to

treat a much wider range of series, such as branch-wise defined series and series

containing several variables.

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Two Fundamental Proofs

Let’s begin with two elementary preliminary statements.

Statement 1

A convergent sequence is bounded.

Proof:

Consider a convergent sequence{an } with limn → ∞

an=Μ ¿ . This means that for every ε>0

there exists an N such that for all n>N ,

|an−Μ ¿|<ε ⇔−ε<an−Μ ¿<ε ⇔ Μ ¿−ε<an<Μ ¿+ε .

Choose ε=Μ ¿

2. The above inequality gives

Μ¿

2<an<

3 Μ ¿

2.

Choose B to be the maximum of the set {a1 , a2 , a3 , …, aN −1,3 Μ ¿

2 } and b to be the

minimum of the set{a1 , a2 , a3 , …, aN −1,Μ ¿

2 }. Then, for every n ≥ 1 ,b≤ an≤ B . Hence,

sequence {an } is bounded.

QED

Statement 2

A monotonic sequence that is bounded converges.

Proof:

Consider, for simplicity, an increasing sequence {an }, which is bounded above.

i.e. for all n∈N , the values a1 , a2 ,…,an are less than M, for some M real and constant.

Of all the numbers M that satisfy this, let Μ ¿ be the smallest (this is called the Least

Upper Bound Axiom.)

Choose any ε>0 and consider the number: Μ ¿−ε . This is clearly not an upper bound of

the sequence, as Μ ¿ is the smallest upper bound that exists. Hence, there exists an N∈N

such that for all n>N , Μ¿−ε<an<Μ ¿<Μ ¿+ε .

Therefore, −ε<an−Μ¿<ε ⇔|an−Μ ¿|<ε, which by definition means that

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limn → ∞

an=Μ ¿. Hence, sequence an converges to Μ ¿ .

QED

Elementary Convergence Tests

The first proposition is called the (direct) comparison test. One version of it goes as follows:

Proposition 1 (Comparison Test)

Suppose sequences {an }n=1

∞ and {bn }n=1

∞ are such that 0 ≤ an ≤ bn for all n ≥ 1.

If ∑n=1

∞

bn converges, so does ∑n=1

∞

an

Furthermore, if 0 ≤ bn ≤ an, then, if ∑n=1

∞

bndiverges, so does ∑n=1

∞

an

Proof

Case I: 0 ≤ an ≤ bn

If Sn=∑k=1

n

ak=a1+a2+…+anand T n=∑k=1

n

bk=b1+b2+…+bnthen 0 ≤ Sn≤ T n for all n.

Now {T n } is bounded, since ∑n=1

∞

bn converges.* Since {Sn } is bounded and monotonic**, it

also converges. Hence, ∑n=1

∞

an converges. QED

Case II:0 ≤ bn ≤ an

If Sn=∑k=1

n

ak=a1+a2+…+an and T n=∑k=1

n

bk=b1+b2+…+bnthen 0 ≤ Tn ≤ Sn for all n.

Now {T n } grows out of bound, since ∑n=1

∞

bn diverges. Therefore {Sn } is getting increasingly

larger, and thus is also unbounded. Hence, Sn=∑n=1

∞

an diverges.

* Statement 1** Statement 2

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QED

Application 1

Consider ∑n=1

∞ 7+sin5 (n+3 )3n+n2 for all n∈N .

Since −1 ≤sin5 (n+3 )≤ 1 ⇔ 6 ≤ 7+sin5 (n+3 )≤ 8. Since 3n+n2>3n>0 (or 1

3n+n2< 1

3n) for all

n∈N , observe that

0 ≤7+sin5 (n+3 )

3n+n2 ≤83n

and ∑n=1

∞83n =8∑

n=1

∞

( 13 )

n

is a convergent geometric series since r=13<1.

Let an=7+sin5 (n+3 )

3n+n2 , bn=8( 13 )

n

. By applying the above proposition, ∑n=1

∞ 7+sin5 (n+3 )3n+n2 is

bounded, and hence converges. QED

We now present a different but equivalent form of the Comparison Test, known as the Limit Comparison Test.

Proposition 2 (Limit Comparison Test)

If sequences {an }n=1

∞ and {bn }n=1

∞ are made up of positive terms and lim

n → ∞ ( an

bn)=c where c>0

(or equivalently limn → ∞ ( an

cbn)=1), then∑an converges if and only if ∑bn converges.

If c=0, the convergence of ∑bn implies that ∑an also converges or equivalently, the

divergence of ∑an implies that ∑bnalso diverges.

If c=∞ the convergence of ∑an implies that ∑bn also converges or equivalently, the

divergence of ∑bn implies that ∑analso diverges.

Proof

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Since limn → ∞ ( an

cbn)=1 ,for all ε>0 there exists an N such that for all n>N ,

| an

cbn

−1|<ε. Let ε=12

. Hence, −12

<an

c ⋅bn

−1< 12⇒ 1

2<

an

c ⋅bn

< 32

. Therefore an<32⋅c ⋅ bn

and cbn

2<an for all n ≥ N . If bn converges, then

32

c ∑bn also converges, and it follows

that ∑ an converges, since it is bounded by the first inequality. If ∑ bn diverges, then

c2

∙∑ bn also diverges, and it follows that ∑ an diverges, since it outgrows

c2

∙∑ bnby the second inequality. QED

Note: The above two propositions are obviously valid even if the sequences are made up eventually of positive terms. i.e. if there exists n0 such that for all n>n0, all the terms an , bn are positive.

Application 2

Consider the sequence an=ln (n )n3+1

. To decide whether it converges or diverges, we

employ the limit comparison test:

Consider bn=

ln (n )n

n2

. The numerator tends to 0 (simple application of de L’Hopital on

ln ( x )x

), so limn → ∞

bn=0.

Consequently, for some N, we know that ln (n )

n≤ 1 for all n ≥ N .

Hence, ln ( n )

n

n2 ≤1

n2

. Since ∑ 1

n2 converges (p-series with p>1), we conclude, by

comparison test, that ∑ ln (n )n3 also converges.

Now, using the limit comparison test:

limn→ ∞

bn

an

=limn→ ∞

ln (n )n3

ln (n )n3+1

=limn → ∞ (1+ 1

n3 )=1≠ 0

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Hence, ∑ an converges.

Our next elementary convergence test is the Integral test.

Proposition 3 (Integral Test)

Suppose sequence {an } is made up of positive terms. Let an=f (n), with f (x) being a

continuous, positive and decreasing function for all x≥ N (where N is a positive integer).

Then, the series ∑n=N

∞

an converges if and only if ∫N

∞

f (x )dx converges.

Proof

Suppose f is decreasing, with f (n)=an for all n.

Observe that in Diagram (a) ∑k =1

n

ak encloses more

area than ∫1

n+ 1

f ( x ) dx does.

That is, ∫1

n+1

f ( x ) dx≤ a1+a2+⋯ an(I )

In Diagram (b), rectangles are drawn under f (x). Disregarding the first term, we

observe that a2+a3+⋯ an ≤∫1

n

f ( x )dx .

Including a1, we get:

a1+a2+a3+⋯ an≤ a1+∫1

n

f ( x ) dx ( II )

Combining (I ) and (II ) gives:

∫1

n+ 1

f ( x ) dx≤ a1+a2+⋯ an≤ a1+∫1

n

f (x ) dx

If ∫1

∞

f ( x )dx is finite, the inequality on the right (through the comparison test) shows

∑n=1

∞

an is finite. Hence ∑ an is convergent

If ∫1

∞

f ( x )dx is not finite, the inequality on the left shows that ∑ an is also infinite (again,

through comparison test). Hence ∑ an is divergent.

QED

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Application 3

As a direct application, consider ∑n=1

∞1

√ x3. In order to apply the Integral Test to the

series, we observe that:

∫1

∞1

x32

dx=[−2 x−12 ]1

∞

=limb → ∞ (−2

1√b

+2)=2

Hence both the integral and the series converge.

Application 4

As a more involved application, consider the following:

Find the values of p for which the following series is convergent: ∑n=2

∞1

n( ln n)p

Let an=1

n( lnn)p . For n ≥ 2, an is obviously positive. Let f ( x )= 1

x (ln x )p . Observe that

f ( n )=an. Furthermore, observe that f ( x ) is positive for x≥ 2.

Also, f' ( x )=−ln p ( x )+p ∙ ln p−1(x )

[ x (ln x)p]2. Clearly, f ' ( x ) is negative for all x. Hence f ( x ) is

decreasing. Therefore, Proposition 3 is applicable, and we deduce that ∑n=2

∞

an has the

same convergence behavior as the integral ∫2

∞

f ( x ). i.e.

∫2

∞1

x (ln x)p dx

Set u = ln x⇒du=dxx

. Hence, for p ≠1(we will treat the case p=1 explicitly at the end)

the integral becomes

∫ln 2

∞1

(u)p du=[ u1−p

1−p ]ln 2

∞

= limM →∞ [ u1−p

1−p ]ln 2

lnM

=1

1−p[ ( lnM )1−p−(ln 2)1−p ] .

For this to converge, we need 1−p<0, otherwise lnM grows indefinitely, thus

diverging. Hence, p>1.

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Now, for p=1, the integral becomes ∫2

∞1

xlnxdx, which upon substitution u=ln (x )

becomes ∫ln 2

∞1u

du=[ ln (u)]ln2

∞, which clearly diverges. Therefore, the above series

converges for p>1.

We now reach the last elementary convergence test treated in the IB Syllabus, namely

the Ratio test.

Proposition 4 (Ratio Test)

Suppose series ∑an with positive terms and suppose that limn → ∞

an+1

an

=λ

(i) If λ<1 then ∑an converges

(ii) If λ>1 then ∑an diverges

(iii) If λ=1, the test is inconclusive

Proof

(i) We suppose that λ<1 and we choose μ so that λ<μ<1

Since ak +1

ak

→ λ, we know there that exists some k o such that if k ≥ ko, then ak +1

ak

<μ

This gives ako+1<μ ⋅ako, ako+2<μ ⋅ako+1<μ2 ⋅ako

, and in a more general form, ako+p<μp ⋅ako,

with p=1 , 2 ,3 ,…

Set p=k−ko

For k>ko, we have ak< μk−ko ⋅ako=

ako

μko

μk (I)

Since 0<μ<1,∑ μk converges

It follow from (I) and the Comparison test that ∑ako

μko

μk=ako

μko

∑ μk converges as well.

QED

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(ii) Suppose that λ>1. This implies that ak+1>ak, and hence the terms will never

approach 0, thus proving divergence of the series. QED

(iii) For λ=1 consider the specific series ∑n=1

∞1n

and ∑n=1

∞1n2

For ∑n=1

∞1n

, applying the ratio test gives an+1

an

=

1n+1

1n

=n

n+1→1

For∑n=1

∞1n2 , applying the ratio test gives

an+1

an

=

1

(n+1)2

1n2

=( nn+1 )

2

→1

Yet, from integral test, ∫1

∞1n=[ ln ( n ) ]1

∞=lim

b → ∞ln (b ) diverges whereas

∫1

∞1n2 =[−1

n ]1

∞

=1 converges.

In conclusion, for λ=1, the Ratio Test is inconclusive and a different test should be used.

QED

We will not supply any examples on the ratio test since this is standard IB Syllabus

option Calculus material.

For a counterexample to the Ratio Test, consider the series ∑ ak, where ak= 1

2k when k

is odd and ak = 1

2k−2 when k is even.

The ratio test cannot deal with this series since if we take the ratio of the even to the odd,

the series diverges (lim a2n

a2 n+1

=2), whereas if we take the ratio of the odd to the even, the

series converges(lim a2 n+1

a2 n

=12).

At this point, we will introduce the first advanced test that is not treated in the syllabus,

namely the Root Test (of which the Ratio Test is a special case). The Root Test is used

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for many series, including some that the Ratio Test fails to classify as convergent or

divergent.

Root Test

Proposition 5 (Root Test)

Suppose series ∑an with an≥ 0 for n ≥ N and suppose that limn → ∞

n√an= λ

Then

(i) The series converges if λ<1

(ii) The series diverges if λ>1

(iii) Test is inconclusive if λ=1

Proof

(i) Suppose λ<1and choose ε>0, such that ε+ λ<1. Since n√an → λ, the terms n√an

eventually get closer to λ than ε . In other words, for some N∈N , n√an< λ+ε when n ≥ N .

It follows that an< ( λ+ε )n for n ≥ N .

Now, ∑n=N

∞

( λ+ε )nis a geometric series with ratio λ+ε<1, hence convergent. Therefore,

∑n=M

∞

an converges, and ∑n=1

∞

an does as well.

QED

(ii) Suppose now that λ>1. For all N∈N , we have n√an>1⇒ αn>1 for n>N . Therefore,

limn → ∞

α n≠ 0, so the series diverges.

QED

(iii) For λ=1 consider ∑n=1

∞1n

and ∑n=1

∞1n2

For ∑n=1

∞1n

, applying the root test gives limn → ∞

n√ 1n=1

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For ∑n=1

∞1n2 , applying the root test gives lim

n → ∞

n√ 1n2=1

As we have already proven, however, the first series diverges, while the second one

converges. Hence, we are led to conclude that for λ=1 the Root Test is inconclusive as

well.

QED

Note: It is important to realize that the Root Test is more general than the Ratio Test in

the sense that all cases decided by the Ratio Test can in principle also be decided by the

Root Test but not the other way around. There exist cases, such as the one treated below,

where the Ratio Test is inconclusive but the Root Test does decide.

Application 5

Again consider the series introduced before, namely ak = 1

2k if k is odd and ak = 1

2k−2 if k

is even. As shown before, the ratio test fails to decide on the convergence of this series.

Applying the root test, however, we can treat the even and odd terms as two sub-series

and be led to the following results:

limn → ∞

n√ a2n+1

a2n−1

=limn → ∞

n√ 1

2n=1

2∧lim

n→ ∞

n√ 1

2n−2=1

2= lim

n →∞

n√ a2n

a2 n−2

Since both limits are less than 1, both sub-series converge and thus the whole series

converges.

(Note that, in reality, the above series is just a rearrangement of the convergent

geometric series ∑n=0

∞12n .)

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Advanced Convergence Tests

We now move to a more advanced class of convergence tests that allow for multiple

parametric dependences of sequences and treat “tail” behavior. The first one is

Kummer’s test, which will later be used in the proofs of Raabe’s and Gauss’s tests.

Kummer's test

Proposition 6 (Kummer’s Test)

Let {an } and {bn } be sequences with positive terms for all n ≥ N , and cn=bn−bn+1 an+1

an.

(i) If there is a positive constant r such that cn ≥r>0 for all n ≥ N then ∑an converges.

(ii) If cn ≤0 for n ≥ N and if ∑1bn

diverges, then ∑an diverges.

Proof of (i):

We are given that cn=bn−bn+1 an+1

an

≥ r>0.

Hence, bn an−bn+1an+1

an

≥ r or bnan−bn+1 an+1≥ r ⋅an

Writing it out explicitly:In general, for any n=K , we have bK aK−bK +1aK +1 ≥r ⋅aK Writing the above relation explicitly for k=N , N+1, N+2…,we find:

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+bN aN−bN +1 aN +1 ≥r ⋅ aN k=N+bN +1 aN+1−bN+2aN +2≥ r ⋅aN +1 k=N+1

…

…+bn an−bn+1 an+1≥ r ⋅an k=n

Executing the partial cancellations, we arrive at:bN aN−bn+1 an+1 ≥ r ⋅ ( aN+aN +1+…+an )

From this, we deduce that ∑k=N

n

ak ≤aN bN−an+1 bn+1

r≤

aN bN

r (since

the terms are positive). However, the value of aN bN

r is fixed (it is a constant). Hence,

limn → ∞

∑k=N

n

ak ≤ constant . Since the series is increasing and bounded above, it converges.

QED

Proof of (ii):

First, observe that cn ≤0 is equivalent to bn an−bn+1an+1

an

≤ 0, or, bnan≤ bn+1 an+1, or after

rearranging,

an

1bn

≤an+1

1bn+1

. This means that { an

1bn

} is an increasing sequence with positive

terms. Hence, for someN∈N , we know that for all n>N ,an>1bn

.

Therefore, for n>N ,∑ an>∑ 1bn

. And since ∑ 1bn

diverges, ∑ an does as well.

QED

Having proven Kummer's test, we can proceed in proving two other very important and

more complex tests that will utilize it and help us treat more complex series.

(add)

∑k=N

n

ak

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Raabe's Test

Proposition 7 (Raabe’s Test)

Suppose series ∑an has all terms positive. If there exists r>0 and N ≥1 such that

an+1

an

≤1−1n− r

n for all n ≥ N , then ∑an converges.

The series ∑an diverges if an+1

an

≥1−1n

for all n ≥ N .

Proof

To prove Raabe’s test, we use Kummer's test and deliberately choose bn+1=n.

We know cn=bn−bn+1 an+1

an

=(n−1 )−nan+1

an

Convergence case:

cn ≥r>0

⇒ bn an−bn+1an+1

an

≥ r

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⇒ (n−1 ) an−n∙an+1 ≥r ∙an

⇒ an (n−r−1 )≥ an+1 n

⇒an+1

an

≤ 1−1n− r

nfor sufficiently largen ' s (n≥ N ) QED

Divergence case:

cn ≤0

⇒bn an−bn+1 an+1

an

≤ 0

⇒ (n−1 ) an−n∙an+1 ≤0

⇒ (n−1 ) an ≤ n∙ an+1

⇒an+1

an

≥ 1−1n

for sufficiently largen ' s (n ≥ N )

Kummer's test states that, in order for ∑an to diverge, ∑1bn

should also diverge.

But clearly: 1bn

= 1n−1

> 1n

Hence ∑1n<∑

1n−1

=∑ 1b

. By integral test (or p-series test), ∑1n

diverges, and by

comparison test, ∑1

n−1= 1

bn does so as well.

Since ∑ 1bn

diverges, it is dominated by ∑ an (Kummer's test). Hence, ∑an diverges.

QED

We now present another equivalent but very useful form of the Raabe’s test, namely the

Limit Form. We will find this limit form extremely useful when we proceed to the

studying of Gauss's test.

Alternate form of Raabe's test (Limit form):

an+1

an

≤1−1n− r

n⇒n(1−

an+1

an)≥ (r+1 ) ,r>0

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limn → ∞ [n (1−

an+1an

)]≥ (r+1 )

limn → ∞ [n (1−

an+1an

)]≤ 1

Notice that, for r=0, both limits are equal to 1. Hence, the limit form is inconclusive.

Therefore, the case r=0 must be examined separately.

Before we move to the Gauss’s Test, let’s consider a simple application of Raabe’s Test.

Application 6

Here is an example of the use of Raabe’s test:

Consider ∑n=1

∞2 ⋅4 ⋯2 n

1 ⋅3⋯ (2 n−1 )⋅ 1

2n+2. Decide on the convergence of the series.

Observe that

an+1

an

=

2 ∙ 4 ∙ ∙∙(2n+2)1∙ 3 ∙∙ ∙(2n+1)

2 ∙ 4 ∙ ∙ ∙2 n1∙3 ∙∙ ∙(2 n−1)

=

2 ∙4 ∙ ∙∙ (2 n+2)1∙3 ∙∙ ∙(2n+1)

2 ∙ 4 ∙ ∙ ∙2n1∙ 3 ∙∙ ∙(2n−1)

∙

12n+4

12n+2

=2 n+22 n+1

∙2 n+22n+4

=4 n2+8n+4

4 n2+10n+4.

Also, note that limn → ∞

4 n2+8 n+44 n2+10 n+4

=1, so the Ratio test fails to provide an answer.

Rn=n(1−an+1

an)=n(1− 4 n2+8n+4

4 n2+10 n+4 )=n( 4 n2+10 n+4−4n2−8n−44 n2+10 n+4 )= 2n2

4n2+10 n+4

But limn → ∞

2 n2

4 n2+10 n+4=1

2.

Hence, r+1=12⇒ r=−1

2≤ 0.

So the series diverges.

We now reach the final and most important convergence test studied here, namely Gauss’s Convergence Test.

Gauss's test

Proposition 8 (Gauss’s Test)

if r ≤ 0, ∑an diverges

if r ≥ 0, ∑an converges

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Suppose series ∑an has positive terms. If there is an N ≥1 and s>1, and an M >0 such

that an+1

an

=1− An

+f (n )nS for n ≥ N , where |f (n )|≤ M for all n (i.e. f ( n ) is bounded for all n

), then ∑an converges if A>1 and diverges if A ≤ 1

Proof

To prove Gauss's test, manipulate an+1

an

=1− An

+f (n )nS ⇔(1−an+1

an)= A

n−

f (n )nS ⇔

n(1−an+1

an)=A−

f (n )ns−1 , s>1

lim ¿n⟶∞ n(1−an+1

an)=A−lim ¿n⟶∞

f (n )ns−1 , s>1¿¿

However, 0=−M limn⟶∞

1

ns−1≤ lim

n⟶∞

f (n )ns−1

≤ M limn⟶∞

1

ns−1=0

Hence, from Squeeze Theorem, limn⟶∞

f (n )ns−1

=0.

So lim ¿n⟶∞ n(1−an+1

an)=A ¿

⇒ lim ¿n⟶∞ n(1−an+1

an)=r+1¿

(i) ∑an converges ifr>0

⇒ A−1>0

⇒ A>1

Therefore ∑an converges if A>1.

QED

(ii) ∑an diverges if r<0

Set A=1+r

000901-0086

⇒ A−1<0

⇒ A<1

Therefore, ∑an diverges if A<1. QED

Finally, the case r=0, (i.e. A=1) is the inconclusive case discussed above that must be

separately examined.

Consider, once again, Kummer's test.

To prove that for A=1 a series diverges according to Gauss's test, we deliberately set

bn+1=n ln ( n ) which knowingly diverges.

cn=bn−bn+1 an+1

an = (n−1) ln (n−1 )−n ln ( n )[1−1

n−

f (n)ns ]

¿ (n−1 ) ln (n−1 )−n ln (n )+ln (n )+¿

f (n)ns−1 ln (n )

At this point we should note that since |f (n )|≤ M ,

−Mlimn→ ∞

ln (n )

ns−1 ≤ limn→ ∞ (f (n ) ln (n )

ns−1 )≤ Mlimn →∞

ln (n )

ns−1 (I )

However, limn →∞

ln (n )

ns−1 =limn→∞

1

(s−1 ) ns−1 (By L' Hopitals rule)=0

Hence, by applying the squeeze theorem to (I ), limn→ ∞ ( f (n ) ln (n )

ns−1 )=0

Back to our original equation, taking the limit gives

limn → ∞

cn=limn → ∞

[n ln( n−1n )−ln( n−1

n )+ f (n ) ln (n )ns−1

]

Unknown

000901-0086

⇒ limn → ∞

cn= limn→ ∞

[ ln(1−1n )

n

−ln(1−1n )+ f (n ) ln ( n )

ns−1 ]

So limn → ∞

cn= ln( 1e )−ln (1 )+0=−1

This means that there exists an N∈N such that, for all n>N , cn<0.

Now let’s prove that ∑1bn

=∑1

n ln (n ) diverges

Applying integral test:

∫2

∞1

x ln x= lim

T → ∞¿¿¿

Since the integral does not converge, ∑1bn

=∑1

n ln (n ) also diverges

Thus, both requirements for divergence by Kummer's test are fulfilled; in other words,

∑an diverges. All in all, this means that, for A=1, Gauss's test shows divergence.

QED

Application 7

As a concrete application of Gauss’s test, and a clear case where the aforementioned test

can be utilized to treat tail behavior in series, consider the Hypergeometric series, first

introduced by Gauss:

∑n=0

∞

an=1+ α ⋅ β1⋅ γ

+α (α+1 )⋅ β ( β+1 )

1 ⋅2 ⋅γ ⋅ (γ+1 )+⋯

where α , β, γ are real and positive integers. Observe that:

an+1

an

=

α (α +1 ) (α+2 ) ⋅⋅⋅ (α +n−1 )⋅ ( a+n ) ⋅ β ( β+1 ) ( β+2 )⋅⋅⋅ (β+n−1 ) ⋅ ( β+n )1 ⋅2⋅3 ⋅⋅⋅ n⋅ (n+1 )⋅ γ ⋅ ( γ+1 ) ⋅ (γ +2 )⋅⋅ ⋅ (γ+n−1 ) ⋅ (γ +n )α ( α+1 ) (α+2 )⋅⋅⋅ ( α+n−1 )⋅ β ( β+1 ) ( β+2 ) ⋅⋅⋅ ( β+n−1 )

1 ⋅2⋅ 3⋅⋅⋅ n⋅ γ ⋅ ( γ+1 ) ⋅ (γ +2 )⋅⋅ ⋅ (γ+n−1 )

⇒an+1

an

=

α (α+1)(α+2)⋅⋅⋅(α +n−1)⋅(a+n)⋅ β( β+1)(β+2)⋅⋅⋅(β+n−1)⋅(β+n)1 ⋅2⋅3 ⋅⋅⋅ n⋅(n+1)⋅ γ ⋅(γ+1)⋅(γ+2)⋅⋅⋅(γ+n−1)⋅(γ +n)α (α +1)(α+2)⋅⋅⋅(α+n−1)⋅ β (β+1)(β+2)⋅⋅⋅(β+n−1)

1 ⋅2⋅3 ⋅⋅⋅ n⋅ γ ⋅(γ+1)⋅(γ+2)⋅⋅⋅(γ+n−1)

=(α +n ) ⋅ ( β+n )(γ +n ) ⋅ (1+n )

=n2+ (α+β ) n+α ⋅ β

n2+ (γ+1 )n+γ

Now lim ¿n⟶∞ n(1−an+1

an)¿

000901-0086

¿ lim ¿n⟶∞ n(1−n2+( α+β ) n+α ⋅ β

n2+( γ+1 ) n+γ )¿

¿ lim ¿n⟶∞ n( n2+(γ +1 ) n+γ−n2−(α+β ) n−α ⋅ β

n2+(γ +1 ) n+γ )¿

¿ lim ¿n⟶∞ n( [ (γ+1 )−(α +β ) ]n+γ−α ⋅ β

n2+ (γ+1 )n+γ )¿¿ lim ¿n⟶∞(

n2 [ (γ+1 )−(α+ β ) ]+γ−α ⋅ β

n2+(γ +1 ) n+γ)¿

¿ lim ¿n⟶∞([ (γ+1 )−(α +β ) ]+ γ

n2−α ⋅ β

n2

1+(γ+1 )

n+ γ

n2

)=( γ+1 )−(α +β)¿

Hence, the series converges for (γ +1 )−( α+β )>1⇒γ >α+ β and diverges for (γ +1 )−( α+β ) ≤1⇒γ ≤ α+βAn interesting as well as useful notation is achieved through the introduction of the

Pochhammer symbol {q }n≡ q ⋅(q+1)⋅(q+2)⋅⋅⋅(q+n−1)for n ≥ 1.

We can now write the hypergeometric series as follows:

∑n=0

∞

an=1+∑n=1

∞

an ,where∑n=1

∞

an=α ⋅ β1 ⋅γ

+α (α +1 )⋅ β ( β+1 )

1⋅2 ⋅γ ⋅ (γ+1 )+⋯

Hence, an=α (α +1 )⋅⋅⋅(α +n−1)⋅ β ( β+1 ) ⋅⋅⋅(β+n−1)

n! ⋅ γ ⋅ ( γ+1 ) ⋅⋅⋅(γ+n−1)=

{a }n ⋅{β }n

n! ⋅ {γ }n.

i.e.

∑n=0

∞

an=1+∑ {a }n ⋅{β }n

n! ⋅ {γ }nType equation here .

Application 8

As an example of the use of the Hypergeometric series investigated above, consider the

series:

∑n=0

∞

an=1+ 157

+ 360112

+ 126003024

+⋯

which if written in the form

∑n=0

∞

an=1+ 3 ⋅51 ⋅7

+ 3 ⋅ 4 ⋅5 ⋅61⋅2 ⋅7 ⋅8

+ 3 ⋅ 4 ⋅5 ⋅5⋅ 6 ⋅71⋅2 ⋅3 ⋅7 ⋅8 ⋅9

+⋯

000901-0086

is the Hypergeometric series treated above with α=3 , β=5∧γ=7. Since 7 ≤ 3+5, the

divergence criterion is met, the series above diverges (not surprisingly, since the terms

seem to increse beyond bound.)

On the other hand, let's also consider the series

∑n=0

∞

an=1+ 35+24

60+ 360

1260+⋯

which if written in the form

∑n=0

∞

an=1+ 1 ⋅31 ⋅5

+ 1 ⋅2⋅3 ⋅ 41 ⋅2 ⋅5⋅ 6

+1 ⋅2 ⋅3 ⋅3 ⋅4 ⋅51⋅2 ⋅3 ⋅5 ⋅6 ⋅7

+⋯

is the Hypergeometric series with α=1 , β=3∧γ=5. Since 5>1+3, the convergence

criterion is met, so the series above converges.

Conclusion

In this work, I have studied in depth the material on series convergence tests

contained in the IB Higher Level option Calculus syllabus and have provided explicit

proofs of them. Furthermore, I have demonstrated their inability to handle certain series,

either of one or of multiple parameters. I then extended the usage of convergence tests to

more advanced tests in order to study series generally untreatable by the IB syllabus. As

such, I stated, proved and demonstrated the use of the Root Test, as a generalization of

the Ratio Test. Even so, more complicated series, especially those defined by more than

one parameters, still remain untreatable by these elementary conventional tests. Hence, I

proceeded in investigating more advanced convergence tests, namely Kummer’s,

Raabe’s, and Gauss’s convergence tests. Furthermore, I provided examples of series

whose convergence can be examined by these advanced tests, and which can certainly

not be treated by the elementary convergence tests included in the IB syllabus. A famous

example employed is the Gauss Hypergeometric series, which can be studied using

Gauss’s convergence test, but not any other.

000901-0086

Total word count: 2979

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