Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}}{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}$

Short answer

The interval of convergence is$(-1,1]$& radius is 1.

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

Take, $a_n=\frac{{(-1)}^n}{n}{x}^n$then.$a_{n+1}=\frac{{(-1)}^{n+1}}{n+1}{x}^{n+1}$

Apply Ratio Test.

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

Which gives

$\begin{array}{c}=\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{n}{n(1+\frac{1}{n})}\right|\\ =\left|x\right|\underset{n\to \infty }{\lim }\left|\frac{1}{(1+\frac{1}{n})}\right|\\ =\left|x\right|\left|\frac{1}{(1+0)}\right|\\ =\left|x\right|\end{array}$

Therefore, the series converges for $\left|x\right|<1$. That is the radius of convergence is.$R=1$

Check convergence at end points

As.$\left|x\right|<1\Rightarrow -1<x<1$Check the convergence of the series at endpoints $x=-1$and.$x=1$ Substitute $x=-1$in$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{x}^n$

$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{x}^n=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{(-1)}^n\\ =\sum _{n=1}^{\infty }\frac{{(-1)}^{2n}}{n}\\ =\sum _{n=1}^{\infty }\frac{1}{n}\end{array}$

Checking convergence/ divergence atx=-1

Choose$a_n=\frac{{(-1)}^{2n}}{n}$and$b_n=\frac{1}{n}$and determine.$\underset{n\to \infty }{\lim }\frac{a_n}{b_n}$

$\begin{array}{c}\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{\frac{{(-1)}^{2n}}{n}}{\frac{1}{n}}\\ =1\end{array}$

By limit Comparison test, the series diverges for.$x=-1$

Check at x=1

Substitute $x=1$in.$\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{x}^n$

$\begin{array}{c}\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{x}^n=\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}{\left(1\right)}^n\\ =\sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\end{array}$

Consider.

$\begin{array}{c}\sum _{n=1}^{\infty }\left|a_n\right|=\sum _{n=1}^{\infty }\left|\frac{{(-1)}^n}{n}\right|\\ =\sum _{n=1}^{\infty }\frac{1}{n}\end{array}$

By Cauchy's convergence condition, the series converges conditionally at$x=1$.

Therefore, the interval of convergence is $(-1,1]$and the radius of convergence is.$R=1$