Proving $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}x^{2n-1}$ is differentiable

Question

Let $f:(-R,R)\to\mathbb{R}$ $$x\mapsto\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}x^{2n-1}$$ where $R$ is the radious of convergence.

Prove that $f(x)$ is differentiable on $(-R,R) $ and $f'(x)=\frac{1}{1+x^2}$

First I found $R$:

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

$$ \lim_{n}\left|\frac{a_{n+1}}{a_n}\right|= \lim_{n}\frac{2n-1}{2n+1}\left|x^2\right|=|x^2|$$

So the series is convergent $\Leftrightarrow \, |x^2|<1\Leftrightarrow x\in (-1,1)\Rightarrow \boldsymbol{R=1} $

Now what should I do to prove $f(x)$ is differentiable? I tried to use the M-test

Answer 1

A general property of power series is this: if $\sum_{n=0}^\infty a_n(x-x_0)^n$ converges on $(x_0-R,x_0+R)$, then its sum is differentiable and\begin{align}\left(\sum_{n=0}^\infty a_n(x-x_0)^n\right)'&=\sum_{n=1}^\infty na_n(x-x_0)^{n-1}\\&=\sum_{n=0}^\infty(n+1)a_{n+1}(x-x_0)^n.\end{align}Applying this to your series, you'll get that\begin{align}\left(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}x^{2n-1}\right)'&=\sum_{n=0}^\infty(-1)^nx^{2n}\\&=\sum_{n=0}^\infty(-x^2)^n\\&=\frac1{1+x^2}.\end{align}