Question

In Problems 1-10 find the interval and radius of convergence for the given power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}$

Short answer

The radius of convergence is$R=\frac{1}{2}$& interval of convergence is$\left(\frac{{\displaystyle -1}}{{\displaystyle 2}},\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)$

Step-by-step solution

Ratio test

The Ratio test for the power series, $\sum _{n=1}^{\infty }{a}_n{(x-a)}^n$is given as,

$\underset{n\to \infty }{lim}\left|\frac{{\displaystyle a_{n+1}{(x-a)}^{n+1}}}{{\displaystyle a_n{(x-a)}^n}}\right|=L$

If, $L<1$then the series converges absolutely.

If, $L>1$then the series diverges.

If, $L=1$then the test is inconclusive.

Apply ratio test

Consider the series $\sum _{n=1}^{\infty }\frac{2^n}{n}{x}^n$.

Here$a_n=\frac{2^n}{n}{x}^n$

So,$\frac{a_{n+1}}{a_n}=\frac{\frac{2^n\cdot 2}{n+1}{x}^n\cdot x}{\frac{2^n}{n}{x}^n}$

Limit of the ratio can be written as,

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{2^n\cdot 2}{n+1}\cdot x^n\cdot x\mid }{\frac{2}{n}\cdot x^n}\right|\\ =\underset{n\to \infty }{\lim }\frac{2n}{n+1}\left|x\right|\\ =\underset{n\to \infty }{\lim }\frac{2}{1+\frac{1}{n}}\left|x\right|\\ =2\left|x\right|\end{array}$

Check convergence of series

The series is convergent for$2\left|x\right|<1$.

i.e. $\left|x\right|<\frac{1}{2}$,$-\frac{1}{2}<x<\frac{1}{2}$

Substitute$x=-\frac{1}{2}$in the series.$\sum _{n=1}^{\infty }\frac{2^n}{n}{x}^n$

The series is.$\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(\frac{-1}{2}\right)}^n=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}$

This series is convergent by alternating series test

Find radius & interval of convergence

Substitute$x=\frac{1}{2}$in the series$\sum _{n=1}^{\infty }\frac{2^n}{n}{x}^n$

The series is.$\sum _{n=1}^{\infty }\frac{2^n}{n}{\left(\frac{1}{2}\right)}^n=\sum _{n=1}^{\infty }\frac{2^n}{n}\cdot \frac{{\left(1\right)}^n}{2^n}=\sum _{n=1}^{\infty }\frac{1}{n}$

This is harmonic series so at$x=\frac{1}{2}$it diverges.

Thus, the radius of convergence is.$R=\frac{1}{2}$

Interval of convergence is.$[\frac{-1}{2},\frac{1}{2})$