Question

How do you find the sum of the infinite geometric series 2-2+2-2+..?

Answer 1

To find the sum of the infinite geometric series 2-2+2-2+… , we have to express it in the form $a+ar+a{{r}^{2}}+a{{r}^{3}}+...+a{{r}^{n-1}}$ , where $a$ is the first term and r is the common ratio. We can represent the given series as $2+2.\left( -1 \right)+2.{{\left( -1 \right)}^{2}}+2.{{\left( -1 \right)}^{3}}+...+2.{{\left( -1 \right)}^{n-1}}$ . If r is in the range $-1 < r < 1$ , then we can find the sum of infinite series as ${{S}_{n}}=\dfrac{a}{1-r},r\ne 1$ . If r is not in this range, we will check whether $\left| r \right|$ is equal to 1. If so, there will be not limit and hence sum, as the sum of the given series diverges.

Complete step by step solution:
We have to find the sum of the infinite geometric series 2-2+2-2+… Let us first see how an infinite geometric series is expressed. Infinite geometric series is given as
$a+ar+a{{r}^{2}}+a{{r}^{3}}+...+a{{r}^{n-1}}$ , where $a$ is the first term and r is the common ratio.
We are given that $2-2+2-2+...$ . Let us represent this series in the form $2+2.\left( -1 \right)+2.{{\left( -1 \right)}^{2}}+2.{{\left( -1 \right)}^{3}}+...+2.{{\left( -1 \right)}^{n-1}}$
From the above equation, we can see that common ratio, $r=-1$ and $a=2$. Let us see if $\left| r \right|$ is greater than, less than or equal to 1.
$\left| r \right|=\left| -1 \right|=1$
We know that if $\left| r \right|=1$ , the series does not converge.
Let us consider a series $\sum\limits_{k=0}^{\infty }{{{a}_{k}}}$ . We can define this series as $\displaystyle \lim_{n \to \infty }\sum\limits_{k=0}^{n}{{{a}_{k}}}$ , where $\sum\limits_{k=0}^{n}{{{a}_{k}}}$ is ${{S}_{n}}$ , which is the ${{n}^{th}}$ partial sum of the series.
We can write $2-2+2-2+...$ as $\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}2}$ . Hence, we can write the sum of $2-2+2-2+...$ as
${{S}_{n}}=\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}2}$
When n is an even number, say 4, let us find ${{S}_{n}}$ .
$\begin{aligned} & {{S}_{n}}=\sum\limits_{k=0}^{4}{{{\left( -1 \right)}^{k}}2} \\\ & ={{\left( -1 \right)}^{0}}.2+{{\left( -1 \right)}^{1}}.2+{{\left( -1 \right)}^{2}}.2+{{\left( -1 \right)}^{3}}.2+{{\left( -1 \right)}^{4}}.2 \\\ & =2-2+2-2+2 \\\ & =2 \\\ \end{aligned}$
Hence, when $n=\text{even}$ , ${{S}_{n}}=\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}2}=2$
Let us find ${{S}_{n}}$ when n is a odd number, say 3.
$\begin{aligned} & {{S}_{n}}=\sum\limits_{k=0}^{3}{{{\left( -1 \right)}^{k}}2} \\\ & ={{\left( -1 \right)}^{0}}.2+{{\left( -1 \right)}^{1}}.2+{{\left( -1 \right)}^{2}}.2+{{\left( -1 \right)}^{3}}.2 \\\ & =2-2+2-2 \\\ & =0 \\\ \end{aligned}$
Hence, when $n=\text{odd}$ , ${{S}_{n}}=\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}2}=0$
We can see that ${{S}_{n}}$ alternates between 2 and 0. We can see that $\displaystyle \lim_{n \to \infty }\sum\limits_{k=0}^{n}{{{\left( -1 \right)}^{k}}2}$ does not converge to any value. This means that, ${{S}_{n}}$ does not converge to any value. There is no limit for the given series.

Note: We can find the sum of infinite series if r is in the range $-1 < r < 1$ .If so, we will be using the formula ${{S}_{n}}=\dfrac{a}{1-r},r\ne 1$ . When two the sum of infinite series oscillates between two values, there will not be a sum. That is, there will not be any limit. This occurs when $\left| r \right|=1$ .