# Convergence of Infinite Series

Sequence and Series / Tuesday, November 5th, 2019

# Infinite Series Introduction

The present article introduces briefly the theory of infinite series. Unlike finite series, the sum of an arbitrary infinite series may not be obtained. When it is obtained the infinite series is named convergent series. We begin with definitions of convergence, divergence, and oscillatory series. Also, we introduce a number of useful tests for the convergence of infinite series.

## Definition of Infinite Series

If {un} i.e., { u1, u2, u3, ….., un, …} is a sequence of real numbers, then the expression

${{u}_{1}}+{{u}_{2}}+{{u}_{3}}+…+{{u}_{n}}+…~~upto~~\infty$

is called an infinite series and is usually denoted by

$\sum\limits_{n=1}^{\infty }{{{u}_{n}}}~~or~~more~~briefly~~\sum{{{u}_{n}}}$

Convergent Series
Divergent series
Oscillatory series
Geometric series
Comparison Test
The hyperharmonic series or the p-series
D’ Alembert’s Ratio Test
Raabe’s Test

## Convergent Infinite Series

A series Σun of real numbers is said to be convergent and converges to u if the sequence {sn} is convergent and lim sn = u where sn = u1 + u2 + … + un, for n = 1, 2, …. Then we write Σun = n.

Note: sn is said to be the n-th partial sum of the series.

 Example

Consider the series

$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+…~~to~~\infty$

Here

${{u}_{n}}=\frac{1}{n\left( n+1 \right)}=\frac{1}{n}-\frac{1}{n+1}$

$\therefore ~~{{u}_{1}}=1-\frac{1}{2}$

$\therefore ~~{{u}_{2}}=\frac{1}{2}-\frac{1}{3}$

$\therefore ~~{{u}_{3}}=\frac{1}{3}-\frac{1}{4}$

$…………………..$

$\therefore ~~{{u}_{n}}=\frac{1}{n}-\frac{1}{n+1}$

$\therefore {{s}_{n}}={{u}_{1}}+{{u}_{2}}+{{u}_{3}}+…+{{u}_{n}}$

$\Rightarrow {{s}_{n}}=\left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+…+\left( \frac{1}{n}-\frac{1}{n+1} \right)$

$\Rightarrow {{s}_{n}}=1-\frac{1}{n+1}$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,{{s}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\left( 1-\frac{1}{n+1} \right)=1-0=1$

Hence the series is convergent and converges to 1.

## Divergent Infinite Series

A series Σun of real numbers is said to be divergent if the sequence {sn} is divergent where sn = u1 + u2 + … + un, for n = 1, 2, ….Then we write Σun = ∞ or –according as lim sn = ∞ or –.

 Example

Consider the series

${{1}^{2}}+{{2}^{2}}+{{3}^{2}}+…~~to~~\infty$

Here

${{u}_{n}}={{n}^{2}}$

$\therefore {{s}_{n}}={{u}_{1}}+{{u}_{2}}+{{u}_{3}}+…+{{u}_{n}}$

$\Rightarrow {{s}_{n}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+…~+{{n}^{2}}$

$\therefore {{s}_{n}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,{{s}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6}=\infty$

Hence the series is divergent and diverges to +∞.

## Oscillatory Infinite series

A series Σun of real numbers is said to be oscillator if the sequence {sn} is oscillatory, where where sn = u1 + u2 + … + un, for n = 1, 2, ….

Note: If a series is neither convergent nor divergent then the series is oscillatory.

 Example

Consider the series 1 – 1 + 1 – 1 + … to ∞.

Here

${{u}_{n}}={{\left( -1 \right)}^{n-1}}$

So the n-th partial sum,

${{s}_{n}}={{u}_{1}}+{{u}_{2}}+{{u}_{3}}+…+{{u}_{n}}$

$\Rightarrow {{s}_{n}}=1-1+1-1+…~~to~~n~~terms$

$\Rightarrow {{s}_{n}}=0~~or~~1,~~according~~as~~n~~is~~even~~or~~odd$

Thus the sequence of partial sums is {1, 0, 1, 0, …}. So the sequence {sn}  oscillates finitely. Hence the series is oscillatory.

## Geometric Infinite series

The series Σ xn = 1 + x + x2 + x3 + … is known as geometric series with common ratio x.

 Theorem

A geometric series Σ xn is convergent if -1 < x < 1, divergent if x ≥ 1 and is oscillatory if x ≤ -1.

 Example

Consider the series

$1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+…~~to~~\infty$

Clearly the series is a geometric series with common ratio ½ (< 1). Hence the series is convergent.

Some properties of an infinite series

If Σun is convergent and Σun  = s, then Σkun is convergent and Σkun = ks, k is a constant.

If Σun is divergent, then Σkun is divergent, where k is a constant.

If Σun and Σvn are convergent and converges to s1 and s2 respectively, then the series Σ (k1un + k2vn) is convergent and converges k1s1 + k2s2, where k1, k2 are constants.

The character of convergence or divergence of an infinite series remains unaltered by the addition or removal of a finite number of its terms.

 Theorem 01

If the series Σun of real number is convergent, then

$\underset{n\to \infty }{\mathop{\lim }}\,~~{{u}_{n}}=0$

Proof:

$Let~~\sum{{{u}_{n}}=s}$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,~~{{s}_{n}}=s$

Where sn is the n-th partial sum of the series.

Obviously then

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,~~{{s}_{n-1}}=s$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,~~\left( {{s}_{n}}-{{s}_{n-1}} \right)=s-s=0$

$Hence~~\underset{n\to \infty }{\mathop{\lim }}\,~~{{u}_{n}}=0.$

Note: The converse of the above theorem is not true.

 Theorem 02

A positive term series either converges or diverges to +∞, not oscillatory.

## Comparison Test

 Theorem 01

Let Σun and Σvn be two series of positive numbers and Σvn is convergent. Then Σun is convergent if

(1) There exists a positive integer N such that un ≤ kvn, for all n ≥ N, where k is a constant.

Or

(2) If l is a finite number (l may be zero)

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{u}_{n}}}{{{v}_{n}}}=l$

 Theorem 02

Let Σun and Σvn be two series of positive numbers and Σvn is divergent. Then Σun is convergent if

(1) There exists a positive integer N such that un ≥ kvn, for all n ≥ N, where k is a constant.

Or

(2) If l is a non-zero number (l may be ∞)

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{u}_{n}}}{{{v}_{n}}}=l$

 Theorem 03

Let Σun and Σvn be two series of positive numbers. Then

(1)

$if~~\frac{{{u}_{n}}}{{{u}_{n+1}}}>\frac{{{v}_{n}}}{{{v}_{n+1}}}$

For all values of n and Σvn is convergent, then the series Σun is also convergent.

(2)

$if~~\frac{{{u}_{n}}}{{{u}_{n+1}}}<\frac{{{v}_{n}}}{{{v}_{n+1}}}$

For all values of n and Σvn is divergent, then the series Σun is also divergent.

 Example 01

Show that the series

$1+\frac{1}{1.3}+\frac{1}{1.3.5}+~…~~to~~\infty ~~is~~convergent.$

Solution:

Let un (n = 1, 2, 3, …) denote the n-th term of the given series.

Then

${{u}_{n}}=\frac{1}{1.3.5…~~to~~n~~number~~of~~factors}$

$=\frac{1}{1.3.5…~~\left( 2n-1 \right)}<\frac{1}{1.2.2…~\left( n-1 \right)~number~~of~~factors}=\frac{1}{{{2}^{n-1}}}$

$Let~~{{v}_{n}}=\frac{1}{{{2}^{n-1}}}$

$\therefore ~~{{u}_{n}}\le {{v}_{n}},~~for~~n=1,2,…$

But the series

$\sum{{{v}_{n}}}=\sum{\frac{1}{{{2}^{n-1}}}}$

$i.e.,~~1+\frac{1}{2}+\frac{1}{{{2}^{2}}}+…$

is convergent. [Since, it is geometric series with common ratio ½ < 1]

Hence by comparison test, the series Σun is convergent.

 Example 02

Test the convergence of the series

$1+\frac{2}{1!}+\frac{{{2}^{2}}}{2!}+\frac{{{2}^{3}}}{3!}+\frac{{{2}^{4}}}{4!}+…~~to~~\infty$

Solution:

Let un (n = 1, 2, 3, …) denote the n-th term of the given series.

Then

${{u}_{n}}=\frac{{{2}^{n-1}}}{\left( n-1 \right)!}$

Now,

${{u}_{n}}=\frac{2}{n-1}.\frac{2}{n-2}.\frac{2}{n-3}…..\frac{2}{4}.\frac{2}{3}.\frac{2}{2}.\frac{2}{1}$

$<\left( \frac{2}{3}.\frac{2}{3}.\frac{2}{3}….\frac{2}{3} \right).\frac{2}{2}.\frac{2}{1}={{\left( \frac{2}{3} \right)}^{n-3}}.1.2,~~for~~n>4$

But the series

$\sum\limits_{n=1}^{\infty }{{{\left( \frac{2}{3} \right)}^{n-3}}}$

is convergent, since it is a G.P. series with common ratio 2/3 < 1. So by comparison test the series i.e., the given series is convergent.

## The Hyper Harmonic Series or the p-Series

The series

$\frac{1}{{{1}^{p}}}+\frac{1}{{{2}^{p}}}+\frac{1}{{{3}^{p}}}+…~~to~~\infty ~~i.e.,~~\sum\limits_{n=1}^{\infty }{\frac{1}{{{n}^{p}}}}$

is called p-series.

 Theorem

The p-series is convergent if p > 1 and divergent if p ≤ 1.

 Example 01

Test the convergence of the series

$\sum\limits_{n=1}^{\infty }{\frac{\sqrt{n}}{{{n}^{2}}+1}}$

Solution:

The nth term of the series

${{u}_{n}}=\frac{\sqrt{n}}{{{n}^{2}}+1}$

Now, in the ratio un, degree of the expression in the denominator – degree of the expression in numerator = 2 – ½ = 3/2. So we consider a series Σ vn where

${{v}_{n}}=\frac{1}{{{n}^{\frac{3}{2}}}}$

Then,

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{u}_{n}}}{{{v}_{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt{n}}{{{n}^{2}}+1}.{{n}^{\frac{3}{2}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{2}}}{{{n}^{2}}+1}$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{1+\frac{1}{{{n}^{2}}}}=\frac{1}{1+0}=1,~~a~~finite~~number.$

Hence by comparison test, Σ un is convergent or divergent if Σ vn converges or diverges. But Σ vn is a convergent series (being a p-series with p = 3/2 > 1). Therefore the series Σ un is convergent.

 Example 02

Test the convergence of the series

$\frac{6}{1.3.5}+\frac{8}{3.5.7}+\frac{10}{5.7.9}+…~~to~~\infty$

Solution:

The nth term of the series

${{u}_{n}}=\frac{2n+4}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 2n+3 \right)}$

We consider the series Σ vn where

${{v}_{n}}=\frac{1}{{{n}^{2}}}$

Since degree of the expression in denominator – that in numerator in un = 3 – 1 = 2.

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{u}_{n}}}{{{v}_{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{n}^{2}}\left( 2n+4 \right)}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 2n+3 \right)}$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\left( 2+\frac{4}{n} \right)}{\left( 2-\frac{1}{n} \right)\left( 2+\frac{1}{n} \right)\left( 2+\frac{3}{n} \right)}=\frac{2+0}{\left( 2-0 \right)\left( 2+0 \right)\left( 2+0 \right)}$

$=\frac{1}{4},~~a~~finite~~number.$

Hence by comparison test, Σ un is convergent or divergent if Σ vn converges or diverges. But Σ vn is a convergent series (being a p-series with p = 2 > 1). Therefore the series Σ un is convergent.

## D’ Alembert’s Ratio Test

 Theorem

Let Σ un be a series of positive numbers such that

$\underset{n\to \infty }{\mathop{\lim }}\,~~\frac{{{u}_{n+1}}}{{{u}_{n}}}~~exists~~finitely~~and~~\underset{n\to \infty }{\mathop{\lim }}\,~~\frac{{{u}_{n+1}}}{{{u}_{n}}}=l.$

Then the series is convergent if l < 1 and divergent if l > 1.

Note: The above test fails if l = 1.

 Example 01

Test the convergence of the series

$\sum\limits_{n=1}^{\infty }{\frac{n!{{2}^{n}}}{{{n}^{n}}}}$

Solution:

Let un (n = 1, 2, 3, …) denote the n-th term of the given series.

Then

${{u}_{n}}=\frac{n!{{2}^{n}}}{{{n}^{n}}}$

$\therefore {{u}_{n+1}}=\frac{\left( n+1 \right)!{{2}^{n+1}}}{{{\left( n+1 \right)}^{n+1}}}$

$\therefore \frac{{{u}_{n+1}}}{{{u}_{n}}}=\frac{\left( n+1 \right)!{{2}^{n+1}}}{{{\left( n+1 \right)}^{n+1}}}.\frac{{{n}^{n}}}{n!{{2}^{n}}}=\frac{2}{{{\left( n+1 \right)}^{n}}}.{{n}^{n}}$

$\Rightarrow \frac{{{u}_{n+1}}}{{{u}_{n}}}=\frac{2}{{{\left( 1+\frac{1}{n} \right)}^{n}}}$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,~\frac{{{u}_{n+1}}}{{{u}_{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,~\frac{2}{{{\left( 1+\frac{1}{n} \right)}^{n}}}=\frac{2}{e}<1~~\left[ \because ~~2<e<3 \right]$

Hence by D’ Alembert’s ratio test the given series is convergent.

## Raabe’s Test

 Theorem

Let Σ un be a series of positive numbers such that

$\underset{n\to \infty }{\mathop{\lim }}\,~~\left\{ n\left( \frac{{{u}_{n}}}{{{u}_{n+1}}}-1 \right) \right\}~~exists~~finitely~~and~~\underset{n\to \infty }{\mathop{\lim }}\,~~\left\{ n\left( \frac{{{u}_{n}}}{{{u}_{n+1}}}-1 \right) \right\}=l.$

Then the series is convergent if l > 1 and divergent if l < 1.

Note: The above test fails if l = 1.

 Example 01

Test the following series

$\frac{\alpha }{\beta }+\frac{1+\alpha }{1+\beta }+\frac{\left( 1+\alpha \right)\left( 2+\alpha \right)}{\left( 1+\beta \right)\left( 2+\beta \right)}+…~~to~~\infty$

Solution:

Omitting the first term, let un (n = 1, 2, 3, …) denote the n-th term of the given series. Then

${{u}_{n}}=\frac{\left( 1+\alpha \right)\left( 2+\alpha \right)….\left( n-1+\alpha \right)}{\left( 1+\beta \right)\left( 2+\beta \right)\left( n-1+\beta \right)}$

$\therefore {{u}_{n+1}}=\frac{\left( 1+\alpha \right)\left( 2+\alpha \right)….\left( n-1+\alpha \right)\left( n+\alpha \right)}{\left( 1+\beta \right)\left( 2+\beta \right)\left( n-1+\beta \right)\left( n+\beta \right)}$

$\therefore \underset{n\to \infty }{\mathop{\lim }}\,~\frac{{{u}_{n+1}}}{{{u}_{n}}}=\underset{n\to \infty }{\mathop{\lim }}\,~\frac{n+\alpha }{n+\beta }=\underset{n\to \infty }{\mathop{\lim }}\,~\frac{1+\frac{\alpha }{n}}{1+\frac{\beta }{n}}=\frac{1+0}{1+0}=1$

Thus the D’ Alembert’s ratio test fails.

Now,

$\underset{n\to \infty }{\mathop{\lim }}\,~~\left\{ n\left( \frac{{{u}_{n}}}{{{u}_{n+1}}}-1 \right) \right\}=\underset{n\to \infty }{\mathop{\lim }}\,~n\left\{ \frac{n+\beta }{n+\alpha }-1 \right\}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{n\left( \beta -\alpha \right)}{n+\alpha }$

$=\underset{n\to \infty }{\mathop{\lim }}\,\frac{\beta -\alpha }{1+\frac{\alpha }{n}}=\frac{\beta -\alpha }{1+0}=\beta -\alpha$

Hence by Raabe’s test, the given series is convergent when β – α > 1 and divergent when β – α < 1.

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