(-1)"(x - 1)2n+I Σ 2n + 1 n=0

Given the series $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n{\left(x-1\right)}^{2n+1}}{2n+1}$.

To find the radius and interval of convergence.

(a) Let us first find the radius of convergence of the series.

Here $a_n=\frac{{\left(-1\right)}^n{\left(x-1\right)}^{2n+1}}{2n+1}$ and $a_{n+1}=\frac{{\left(-1\right)}^{n+1}{\left(x-1\right)}^{2n+3}}{2n+3}$

We know the ratio test for convergence, for the series $\sum _{n=0}^{\infty }{a}_n$ series converges absolutely if $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|<1$ otherwise diverges.

$\Rightarrow \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{{\left(-1\right)}^{n+1}{\left(x-1\right)}^{2n}\times {\left(x-1\right)}^3}{2n+3}\times \frac{2n+1}{{\left(-1\right)}^n{\left(x-1\right)}^{2n}\left(x-1\right)}\right|$

$\Rightarrow \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|{\left(x-1\right)}^2\right|\underset{n\to \infty }{\lim }\left|\frac{2n+1}{2n+3}\right|$

$\Rightarrow \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|{\left(x-1\right)}^2\right|\left(1\right)$

$\Rightarrow \underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|={\left(x-1\right)}^2$

For convergence ${\left(x-1\right)}^2<1$.

Therefore radius of convergence is $1$.

Now to find interval of convergence let us solve ${\left(x-1\right)}^2<1.$

$\Rightarrow -1<x-1<1$

$\Rightarrow 0<x<2$

Now let us check for endpoints.

For $x=0$ the series will becomes $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{3n+1}}{2n+1}$.

This is an alternating series with $a_n=\frac{1}{2n+1}$.

Here $a_{n+1}<a_n$.

We know the alternating series test the series $\sum _{n=0}^{\infty }{a}_n$  converges if $a_n>0, a_n$ is decreasing and $\underset{n\to \infty }{\lim }{a}_n=0.$

Now $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{2n+1}=0$.

By alternating series test the series $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{3n+1}}{2n+1}$ converges.

Now for $x=2$ the series will becomes $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{2n+1}$.

This is an alternating series with $a_n=\frac{1}{2n+1}$.

Here $a_{n+1}<a_n$.

Now $\underset{n\to \infty }{\lim }{a}_n=\underset{n\to \infty }{\lim }\frac{1}{2n+1}=0$.

By alternating series test the series $\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n}{2n+1}$ converges.

Therefore the interval of convergence of the series $0\le x\le 2.$