Find the interval of convergence, including end-point tests :$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}}^{\mathbf{n}}{\mathbf{x}}^{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}}{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}$

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

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

Then the series$\sum _{n=0}^{\infty }{a}_n$ is convergent if $\rho <1$, divergent if $\rho >1$, and use another set if$\rho =1$ .

For the endpoints test we use the following theorem:

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

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

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

$$\begin{gathered} {\rho }_n=\left|\frac{a_{n+1}}{a_n}\right| \\ =\left|\frac{\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)}\mid }{\frac{{(-1)}^n{x}_{n-1}}{(2n-1)}}\right| \\ =\left|\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)}·\frac{(2n-1)}{{(-1)}^n{x}^{2n-1}}\right| \\ =\left|(-1)·x^2\right| \\ =\left|x^2·\frac{2n-1}{2n+1}\right| \\ \rho =li{m}_{n\to \infty }\left({\rho }_n\right)=li{m}_{n\to \infty }\left|x^2·\frac{2n-1}{2n+1}\right|=\left|x^2\right|. \end{gathered}$$

The series is convergent once

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

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

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

$\begin{array}{rcl}\sum _{n=1}^{\infty }\frac{{(-1)}^n{(-1)}^{2n-1}}{2n-1}& =& \sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}}{2n-1}\\ & =& \frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\cdots \frac{{(-1)}^{n+1}}{2n-1}\cdots .\\ & & \end{array}$

Also,$li{m}_{n\to \infty }\frac{1}{2n-1}=0$ .

Hence, the series$\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}}{2n-1}=\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\cdots \frac{{(-1)}^{n+1}}{2n-1}\cdots$

converges.

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

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

.$\begin{array}{rcl}\sum _{n=1}^{\infty }\frac{{(-1)}^n{\left(1\right)}^{2n-1}}{2n-1}& =& \sum _{n=1}^{\infty }\frac{{(-1)}^n}{2n-1}\\ & =& -\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\cdots \frac{{(-1)}^{n+1}}{2n-1}\cdots \right)\\ & & \end{array}$

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 $-1⩽x⩽1$.