# 6.1 Power series and functions  (Page 2/8)

 Page 2 / 8
$|{c}_{n}{x}^{n}|=|{c}_{n}{d}^{n}|{|\frac{x}{d}|}^{n},$

we conclude that, for all $n\ge N,$

$|{c}_{n}{x}^{n}|\le {|\frac{x}{d}|}^{n}.$

The series

$\sum _{n=N}^{\infty }{|\frac{x}{d}|}^{n}$

is a geometric series that converges if $|\frac{x}{d}|<1.$ Therefore, by the comparison test, we conclude that $\sum _{n=N}^{\infty }{c}_{n}{x}^{n}$ also converges for $|x|<|d|.$ Since we can add a finite number of terms to a convergent series, we conclude that $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ converges for $|x|<|d|.$

With this result, we can now prove the theorem. Consider the series

$\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$

and let S be the set of real numbers for which the series converges. Suppose that the set $S=\left\{0\right\}.$ Then the series falls under case i. Suppose that the set S is the set of all real numbers. Then the series falls under case ii. Suppose that $S\ne \left\{0\right\}$ and S is not the set of real numbers. Then there exists a real number $x*\ne 0$ such that the series does not converge. Thus, the series cannot converge for any x such that $|x|>|x*|.$ Therefore, the set S must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound R . Since $S\ne \left\{0\right\},$ the number $R>0.$ Therefore, the series converges for all x such that $|x| and the series falls into case iii.

If a series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ falls into case iii. of [link] , then the series converges for all x such that $|x-a| for some $R>0,$ and diverges for all x such that $|x-a|>R.$ The series may converge or diverge at the values x where $|x-a|=R.$ The set of values x for which the series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}$ converges is known as the interval of convergence    . Since the series diverges for all values x where $|x-a|>R,$ the length of the interval is 2 R , and therefore, the radius of the interval is R . The value R is called the radius of convergence    . For example, since the series $\sum _{n=0}^{\infty }{x}^{n}$ converges for all values x in the interval $\left(-1,1\right)$ and diverges for all values x such that $|x|\ge 1,$ the interval of convergence of this series is $\left(-1,1\right).$ Since the length of the interval is 2, the radius of convergence is 1.

## Definition

Consider the power series $\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}.$ The set of real numbers x where the series converges is the interval of convergence. If there exists a real number $R>0$ such that the series converges for $|x-a| and diverges for $|x-a|>R,$ then R is the radius of convergence. If the series converges only at $x=a,$ we say the radius of convergence is $R=0.$ If the series converges for all real numbers x , we say the radius of convergence is $R=\infty$ ( [link] ).

To determine the interval of convergence for a power series, we typically apply the ratio test. In [link] , we show the three different possibilities illustrated in [link] .

## Finding the interval and radius of convergence

For each of the following series, find the interval and radius of convergence.

1. $\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$
2. $\sum _{n=0}^{\infty }n\text{!}{x}^{n}$
3. $\sum _{n=0}^{\infty }\frac{{\left(x-2\right)}^{n}}{\left(n+1\right){3}^{n}}$
1. To check for convergence, apply the ratio test. We have
$\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}}{\frac{{x}^{n}}{n\text{!}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)\text{!}}·\frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{\left(n+1\right)·n\text{!}}·\frac{n\text{!}}{{x}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{x}{n+1}|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\frac{1}{n+1}\hfill \\ & =0<1\hfill \end{array}$

for all values of x . Therefore, the series converges for all real numbers x . The interval of convergence is $\left(\text{−}\infty ,\infty \right)$ and the radius of convergence is $R=\infty .$
2. Apply the ratio test. For $x\ne 0,$ we see that
$\begin{array}{cc}\hfill \rho & \hfill =\underset{n\to \infty }{\text{lim}}|\frac{\left(n+1\right)\text{!}{x}^{n+1}}{n\text{!}{x}^{n}}|\\ & =\underset{n\to \infty }{\text{lim}}|\left(n+1\right)x|\hfill \\ & =|x|\underset{n\to \infty }{\text{lim}}\left(n+1\right)\hfill \\ & =\infty .\hfill \end{array}$

Therefore, the series diverges for all $x\ne 0.$ Since the series is centered at $x=0,$ it must converge there, so the series converges only for $x\ne 0.$ The interval of convergence is the single value $x=0$ and the radius of convergence is $R=0.$
3. In order to apply the ratio test, consider
$\begin{array}{cc}\hfill \rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{\left(x-2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}}{\frac{{\left(x-2\right)}^{n}}{\left(n+1\right){3}^{n}}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{\left(x-2\right)}^{n+1}}{\left(n+2\right){3}^{n+1}}·\frac{\left(n+1\right){3}^{n}}{{\left(x-2\right)}^{n}}|\hfill \\ & =\underset{n\to \infty }{\text{lim}}|\frac{\left(x-2\right)\left(n+1\right)}{3\left(n+2\right)}|\hfill \\ & =\frac{|x-2|}{3}.\hfill \end{array}$

The ratio $\rho <1$ if $|x-2|<3.$ Since $|x-2|<3$ implies that $-3 the series converges absolutely if $-1 The ratio $\rho >1$ if $|x-2|>3.$ Therefore, the series diverges if $x<-1$ or $x>5.$ The ratio test is inconclusive if $\rho =1.$ The ratio $\rho =1$ if and only if $x=-1$ or $x=5.$ We need to test these values of x separately. For $x=-1,$ the series is given by
$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}}{n+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\text{⋯}.$

Since this is the alternating harmonic series, it converges. Thus, the series converges at $x=-1.$ For $x=5,$ the series is given by
$\sum _{n=0}^{\infty }\frac{1}{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\text{⋯}.$

This is the harmonic series, which is divergent. Therefore, the power series diverges at $x=5.$ We conclude that the interval of convergence is $\left[-1,5\right)$ and the radius of convergence is $R=3.$

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