# Power Series and Radius of Convergence

You are familiar with different types of series such as arithmetic series , geometric series etc.

As its literal meaning, a series is said to be convergent if it approaches a particular value.

More formally a series is said to be convergent if the sequence of its partial sums approaches a limit as $n$ tends to infinity.

Similarly if the sequence of partial sums does not have a limit or has a limit equal to plus or minus infinity, then the series is divergent.

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Example 1:
**

Determine whether the series $\underset{n=0}{\overset{\infty}{\sum}}{\left(-1\right)}^{n}$ is convergent or divergent.

Consider the first few partial sums.

${s}_{0}=1$

${s}_{1}=1-1=0$

${s}_{2}=1-1+1=1$

${s}_{3}=1-1+1-1=0$

That is, the sequence of partial sums is $\left\{1,0,1,0,1,\mathrm{...}\right\}$ which does not have a limit. So, the sequence is divergent.

Therefore, the series $\underset{n=0}{\overset{\infty}{\sum}}{\left(-1\right)}^{n}$ is divergent.

A
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power series
**
is a series of the form
$\underset{n=0}{\overset{\infty}{\sum}}{c}_{n}{\left(x-a\right)}^{n}$
where
$a$
and
${c}_{n}$
are numbers. The
${c}_{n}$
's are often called the coefficients of the series.

For example, the series $1+2{x}^{2}+3{x}^{3}+4{x}^{4}+\mathrm{....}$ is a power series centered at zero.

A power series is a function of $x$ and therefore whether it converges or diverges also depends upon the values of $x$ . A series may converge for some values of $x$ and may diverge for some other values.

It is guaranteed that a power series about $a$ converges for $x=a$ . You can see that the above power series, $1+2{x}^{2}+3{x}^{3}+4{x}^{4}+\mathrm{....}$ converges to $1$ for $x=0$ .

The radius of convergence is the radius of a large disk in which the series converges.

The biggest interval that a power series converges is called the interval of convergence and this interval is centered at the center of the power series. Note that half of this interval would be equal to the radius of convergence. This could be a non-negative real number or even infinity.

If $r$ is the radius of convergence of a power series centered at $a$ , then it converges for the values of $x$ such that $\left|x-a\right|<r$ and diverges for $\left|x-a\right|>r$ .

That is, the series converges in the interval $a-r<x<a+r$ and diverges for $x<a-r$ and $x>a+r$ .

We can use either ratio test or root test to find the radius of convergence of a power series.

Using the ratio test, a series $\underset{n=0}{\overset{\infty}{\sum}}{c}_{n}{\left(x-a\right)}^{n}$ converges for $x$ such that $\underset{n\to \infty}{\mathrm{lim}}\left|\frac{{C}_{n+1}{\left(x-a\right)}^{n+1}}{{C}_{n}{\left(x-a\right)}^{n}}\right|1$ .

Using the root test, a series $\underset{n=0}{\overset{\infty}{\sum}}{c}_{n}{\left(x-a\right)}^{n}$ converges for $x$ such that $\underset{n\to \infty}{\mathrm{lim}}\left|{\left({C}_{n}{\left(x-a\right)}^{n}\right)}^{\frac{1}{n}}\right|1$ .

**
Example 2:
**

Find the radius of convergence of the power series $\underset{n=0}{\overset{\infty}{\sum}}\frac{{\left(x-1\right)}^{n}}{{2}^{n}}$ .

The given series is a power series centered at $x=1$ . Use the ratio test to find the interval of convergence.

$L=\left|\frac{\left(\frac{{\left(x-1\right)}^{n+1}}{{2}^{n+1}}\right)}{\left(\frac{{\left(x-1\right)}^{n}}{{2}^{n}}\right)}\right|$

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\frac{{\left(x-1\right)}^{n+1}}{{2}^{n+1}}\cdot \frac{{2}^{n}}{{\left(x-1\right)}^{n}}\right|$

$\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left|\frac{x-1}{2}\right|$

Taking the limits, $\underset{n\to \infty}{\mathrm{lim}}\left|\frac{x-1}{2}\right|=\left|\frac{x-1}{2}\right|$ .

The series converges for $x$ such that $\left|\frac{x-1}{2}\right|<1$ .

$\left|\frac{x-1}{2}\right|<1\Rightarrow -1<\frac{x-1}{2}<1$

$\Rightarrow -2<x-1<2$

$\Rightarrow \left|x-1\right|<2$

Therefore, the radius of convergence is $2$ and the interval of convergence is $\left(-1,3\right)$ .