91Ó°ÊÓ

To find the interval of convergence of a given series, we use the ratio test stated below:

Let${\mathrm{ÒÏ}}_n$ be the ratio of two successive terms of an infinite series${\mathrm{ÒÏ}}_n=\frac{a_{n+1}}{a_n}$ and $\mathrm{ÒÏ}=li{m}_{n→∞}\left({\mathrm{ÒÏ}}_n\right)=li{m}_{n→∞}\left(\frac{a_{n+1}}{a_n}\right)$.

Then the series$\underset{n=0}{\overset{∞}{∑}}{a}_n$ is convergent if $\mathrm{ÒÏ}<1$, divergent if $\mathrm{ÒÏ}>1$, and use another set if$\mathrm{ÒÏ}=1$ .

For the endpoints test we use the following theorem:

An infinite series$\underset{n=1}{\overset{∞}{∑}}{(-1)}^n{b}_n$in which the terms are alternately positive and negative is convergent if each term is numerically less than the preceding term and$li{m}_{n→∞}{a}_n=0$.

The given series is $\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n{x}^{2n-1}}{2n-1}$.

Thus, we have$a_n=\frac{{(-1)}^n{x}^{2n-1}}{(2n-1)}$ and $a_{n+1}=\frac{{(-1)}^{n+1}{x}^{2(n+1)-1}}{\left(2\right(n+1)-1)}=\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)}$.

Therefore, the ratio

$$\begin{gathered} {\mathrm{ÒÏ}}_n=\left|\frac{a_{n+1}}{a_n}\right| \\ =\left|\frac{\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)}âˆ\pounds }{\frac{{(-1)}^n{x}_{n-1}}{(2n-1)}}\right| \\ =\left|\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)}·\frac{(2n-1)}{{(-1)}^n{x}^{2n-1}}\right| \\ =\left|(-1)·x^2\right| \\ =\left|x^2·\frac{2n-1}{2n+1}\right| \\ \mathrm{ÒÏ}=li{m}_{n→∞}\left({\mathrm{ÒÏ}}_n\right)=li{m}_{n→∞}\left|x^2·\frac{2n-1}{2n+1}\right|=\left|x^2\right|. \end{gathered}$$

The series is convergent once

$$\begin{gathered} \mathrm{ÒÏ}<1 \\ \left|x^2\right|<1 \\ -1<x<1 \end{gathered}$$.

Thus, the given series is convergent in the interval $-1<x<1$.

At the endpoint $x=-1$, the series is,

$\begin{array}{rcl}\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n{(-1)}^{2n-1}}{2n-1}& =& \underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^{n+1}}{2n-1}\\ & =& \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-⋯\frac{{(-1)}^{n+1}}{2n-1}⋯.\\ & & \end{array}$

Also,$li{m}_{n→∞}\frac{1}{2n-1}=0$ .

Hence, the series$\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^{n+1}}{2n-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-⋯\frac{{(-1)}^{n+1}}{2n-1}⋯$

converges.

Thus, at the endpoint, $x=-1$the given series is convergent.

At the endpoint, $x=1$the series is,

.$\begin{array}{rcl}\underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n{\left(1\right)}^{2n-1}}{2n-1}& =& \underset{n=1}{\overset{∞}{∑}}\frac{{(-1)}^n}{2n-1}\\ & =& -\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-⋯\frac{{(-1)}^{n+1}}{2n-1}⋯\right)\\ & & \end{array}$

Which is the negative of the previous series; hence this series is also convergent.

Thus, at the endpoint,$x=1$ the given series is convergent.

Hence, the given series is convergent in the interval $-1â\copyright ½xâ\copyright ½1$.