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Journal for Studies in Management and Planning

Available at http://edupediapublications.org/journals/index.php/JSMaP/

e-ISSN: 2395-0463

Volume 01 Issue 11

December 2015

Available online: http://edupediapublications.org/journals/index.php/JSMaP/ P a g e | 329

Different Tests to Analyze the Behavior of

Infinite Series - A Review

Joginder Kaur

Lecturer (Mathematics), Punjab Technical University, Punjab, India

ABSTRACT: This paper reviews the methods to

select correct tests for checking the convergence or

divergence of infinite series. To check the

convergence or divergence of a series, there are

various methods like Power Test, Comparison Test,

Root Test, Ratio Test, Gauss Test etc. and the

conditions and methods to apply these tests are

explained here. Only the correct use of these tests

gives us the valid results about the behavior of a

series.

KEY WORDS: Convergent, Divergent, Oscillates,

Partial Sums, Alternating Series, Harmonic Series,

Geometric series, Power Series.

I. INTRODUCTION:

An infinite series is a sequence of numbers in which

an infinite number of terms are added successively.

It is a series in which the number of terms increases

without bound. Of particular concern with infinite

series is whether they are convergent or divergent.

For example, the infinite series 1+1+1+1+-------. is

clearly divergent because the sum of the first n terms

increases without bound as more and more terms are

taken. [9] It is less clear as to whether the harmonic

and alternating harmonic series:

1+1

+ 1

+ ----------

+ 1

+ ----------

Converge or diverge. Indeed one may be surprised to

find that the first is divergent and the second is

convergent. What we shall do here is to consider

some simple convergence tests for infinite series and

explain their applications under different conditions.

[8]

1.1 SEQUENCE OF n

th PARTIAL SUMS:

Before studying the behavior of a series, we shall

know about the sequence of n

th partial sums:

Let ∑ an be an infinite series with partial sums as

σ1= a1

σ2= a1+ a2

σ3= a1+ a2 +a3

---------------------

Then {σn } is called sequence of n

th partial sums.

1.2 BEHAVIOUR OF A SERIES:

Let ∑ an be an infinite series and {σn } be sequence

of n

th partial sums of ∑ an then

(i) ∑ an is convergent iff {σn} is convergent

Further ∑ an = l iff {σn} →l

i.e. Sum of series ∑ an = Limit of sequence {σn}

(ii) ∑ an Diverges to ± ∞ iff {σn} diverges to ± ∞.

(iii) ∑ an Oscillates finitely (or infinitely) iff {σn}

oscillates finitely (or infinitely). [2]

II. DIFFERENT TESTS FOR CHECKING

CONVERGENCE OF AN INFINITE

SERIES:

There are four types of series:

1. Alternating series

2. Power series

3. Geometric series

4. Positive term series

The following table shows different tests used for

different types of series:

Table 1: Different tests for different types of series:

Sr.

No.

Type of series Name of Test

1. Alternating series Leibnitz Test

2. Power series Power test

3. Geometric series Geometric Series Test

4. Positive term series Comparison Test

Cauchy’s Condensation Test

Cauchy’s Root Test

Cauchy’s Integral Formula

D’Alembert’s Ratio Test

Raabe’s Test

Gauss Test

Logarithmic Test

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Journal for Studies in Management and Planning

Available at http://edupediapublications.org/journals/index.php/JSMaP/

e-ISSN: 2395-0463

Volume 01 Issue 11

December 2015

Available online: http://edupediapublications.org/journals/index.php/JSMaP/ P a g e | 330

2.1 LEIBNITZ TEST:

2.1.1 Applicability: It is also known as alternating

series test because with the help of this test, we

can check the convergence of an Alternating

series. [1]

2.1.2 Conditions to Apply and Results:

If the sequence {vn} is

(i) Monotonic

(ii) {vn} →0

Then ∑(−1)

n−1 vn is convergent

EXAMPLE 1. : The series ∑

(−1)

n−1

is convergent.

Here vn =

Clearly (i) { 1

} is monotonically decreasing sequence.

(ii) 1

→ 0 as n→ ∞

So by Leibnitz Test, ∑

(−1)

n−1

is convergent.

2.2 POWER TEST (P- TEST):

2.2.1 Applicability: 1. This test is only used when

the given series is of the form ∑

and it is useful in

various Powerful tests like Comparison test, Caucy’s

Condensation Test etc. [5]

2.2.2 Conditions to Apply and Results: ∑

divergent for p≤1

Convergent for p>1

EXAMPLE 2: The series 1

⁄

is divergent

Because here p= 1

2.3 GEOMETRIC SERIES TEST:

2.3.1 Applicability: This test is used only for

Geometric series in which common ratio of two

consecutive terms is equal. [3]

2.3.2 Conditions to Apply and Results:

Let the given series be of the form 1+x+x

2+x

3+----

Then it converges for |x| <1

And diverges for x ≥ 1

2.4 COMPARISON TEST:

2.4.1 Applicability: 1. This test is used only when

the given series can be broken in to two parts un and

vn and behavior of vn can be easily checked most

probably with the help of Power Test and Geometric

Test.

2.4.2 Conditions to Apply and Results: Let ∑ un

and ∑ vn be two series and limn→∞

= l

(i) If l ≠ 0, l is finite then ∑ un and ∑ vn behave alike.

(ii) If l = 0 then ∑ un is convergent if ∑ vn is convergent.

(iii) If l = infinite then ∑ un is divergent if ∑ vn is

divergent.

EXAMPLE 3: The series ∑

n2+ n+1

is divergent.

Here un=

n2+ n+1

&vn=

Degree of denominator−Degree of numerator

limn→∞

= limn→∞

n2+ n+1

x n

= limn→∞

n2+ n+1

=limn→∞

= 1 ≠ 0

Hence ∑ un and ∑ vn behave alike.

Now ∑ vn = ∑

is divergent. (By P-Test)

Therefore ∑ un = ∑

n2+ n+1

is also divergent.

2.5 CAUCHY’S CONDENSATION TEST:

2.5.1 Applicability: This test is only used when we

know the monotonic behavior of a series. If the series

is monotonically decreasing, then apply Cauchy

Condensation test. [6]

2.5.2 Conditions to Apply and Results: Let ∑ an be

a series such that

(i) an>0 ∀ n

(ii) an > an+1

Then ∑ an and ∑ 2

n a2

n behave alike.

EXAMPLE 4: The series ∑

(log n)

−3

⁄

is divergent.

Here ∑ an= ∑

(log n)

−3

⁄

(i) Clearly an=

n(log n)

⁄

>0 ∀ n

(ii) 1

(n+1)[log( n+1)]

⁄

n(log n)

⁄

∴ an+1 < an

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Journal for Studies in Management and Planning

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e-ISSN: 2395-0463

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December 2015

Available online: http://edupediapublications.org/journals/index.php/JSMaP/ P a g e | 331

Hence ∑ an and ∑ 2

n a2

n behave alike.

∴ ∑ 2

n a2

n = ∑ 2

n 1

n(log 2

⁄

= ∑

(n log 2)

⁄

(log 2)

⁄

∑

⁄

which is divergent by

P-Test (Here p=3

<1)

Hence ∑

(log n)

−3

⁄

is divergent.

2.6 CAUCHY’S ROOT TEST:

2.6.1 Applicability: This test is used when an is

having powers like n, n

etc.

2.6.2 Conditions to Apply and Results: Let ∑ an be

a series such that

(i) an ≥ 0 ∀ n

(ii) limn→∞

1⁄n = l

Then ∑ an is convergent if l < 1

Divergent if l > 1

If l = 1 then Cauchy’s Root Test fails. [4]

EXAMPLE 5: The series ∑(

n+1

)

is convergent

Here (i) an = (

n+1

)

≥ 0 ∀ n

(ii) limn→∞

1⁄n = limn→∞

(

n+1

)

= limn→∞

(

)

< 1

Hence ∑(

n+1

)

is convergent.

2.7. CAUCHY’S INTEGRAL FORMULA:

2.7.1 Applicability: 1. When limits are given i.e. we

have to prove that the given series lies between two

limits then Cauchy’s Integral Test is applied.

2. When limits are not given between which we have

to prove that the given series lies but ∫ f(x) dx is

very easy to calculate then we apply Cauchy’s

Integral Test.

2.7.2 Conditions to Apply and Results: Let f be

defined as non-negative and decreasing ∀ n≥ 1.

Let (i) σn= f (1) +f (2) +f (3) +----------+f (n)

(ii) And In= ∫ f(x) dx n

Then { σn − In} is convergent and

∑ f(n)and limn→∞

∫ f(x) dx n

behave alike.

And if ∑ f(n) is convergent then it converges

between I and I + f (1), where I= limn→∞

2.8 D’ALEMBERT’S RATIO TEST:

2.8.1 Applicability: If an contains factorial of any

term, then we apply Ratio Test.

2.8.2 Conditions to Apply and Results: Let ∑ an be

a series such that

(i) an>0 ∀ n

(ii) limn→∞

an+1

= l

Then ∑ an is convergent if l > 1

Divergent if l < 1

If l = 1 then Ratio Test fails.

EXAMPLE 6: The series ∑

(n!)

2n!

; x > 0 is

convergent for x < 4 and divergent for x > 4.

Here (i) an =

(n!)

2n!

> 0 ∀ n

(ii) an+1 =

(n+1!)

2(n+1)!

n+1

limn→∞

an+1

= limn→∞

(n!)

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

(2n)!(n+1)

2 (n!)

= limn→∞

4n(1+

)

n(1+

)

Hence ∑ an is convergent if 4

> 1

Divergent if 4

< 1

i.e. ∑ an is convergent if x < 4

Divergent if x > 4

And for x=4, Ratio Test fails.

2.9 RAABE’S TEST:

2.9.1 Applicability: 1. When we apply Ratio Test but

it fails and conditions to apply the most common tests

i.e. Gauss Test and Logarithmic Test are not satisfied

then we apply Raabe’sTest.

2.9.2 Conditions to Apply and Results: Let ∑ an be

a series such that

(i) an>0 ∀ n

(ii) limn→∞

(

an+1

−1)n = l

Then ∑ an is convergent if l > 1

Divergent if l < 1

EXAMPLE 7: In example 6, if x=4 then Ratio Test

fails