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 {u_n} i.e., { u₁, u₂, u₃, ….., u_n, …} 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

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

Table of Contents

In this article we will briefly discuss about
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 Σu_n of real numbers is said to be convergent and converges to u if the sequence {s_n} is convergent and lim s_n = u where s_n = u₁ + u₂ + … + u_n, for n = 1, 2, …. Then we write Σu_n = n.

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

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

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 Σu_n of real numbers is said to be oscillator if the sequence {s_n} is oscillatory, where where s_n = u₁ + u₂ + … + u_n, for n = 1, 2, ….

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

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 {s_n}  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.

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

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 Σu_n is convergent and Σu_n  = s, then Σku_n is convergent and Σku_n = ks, k is a constant.

If Σu_n is divergent, then Σku_n is divergent, where k is a constant.

If Σu_n and Σv_n are convergent and converges to s₁ and s₂ respectively, then the series Σ (k₁u_n + k₂v_n) 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.

If the series Σu_n 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 s_n 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.

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

Comparison Test

Let Σu_n and Σv_n be two series of positive numbers and Σv_n is convergent. Then Σu_n is convergent if

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

Or

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

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

Let Σu_n and Σv_n be two series of positive numbers and Σv_n is divergent. Then Σu_n is convergent if

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

Or

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

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

Let Σu_n and Σv_n 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 Σv_n is convergent, then the series Σu_n is also convergent.

(2)

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

For all values of n and Σv_n is divergent, then the series Σu_n is also divergent.

Show that the series

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

Solution:

Let u_n (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 Σu_n is convergent.

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 u_n (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.

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

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 u_n, degree of the expression in the denominator – degree of the expression in numerator = 2 – ½ = 3/2. So we consider a series Σ v_n 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 u_n is convergent or divergent if Σ v_n converges or diverges. But Σ v_n is a convergent series (being a p-series with p = 3/2 > 1). Therefore the series Σ u_n is convergent.

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 v_n where

\[{{v}_{n}}=\frac{1}{{{n}^{2}}}\]

Since degree of the expression in denominator – that in numerator in u_n = 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 u_n is convergent or divergent if Σ v_n converges or diverges. But Σ v_n is a convergent series (being a p-series with p = 2 > 1). Therefore the series Σ u_n is convergent.

D’ Alembert’s Ratio Test

Let Σ u_n 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.

Test the convergence of the series

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

Solution:

Let u_n (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

Let Σ u_n 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.

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 u_n (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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