Continuity, Differentiability, and Integrablity of a Series of Functions I'm working on a problem where I need to show that the series of functions, 
$$
f(x) = \sum_{n\geq 1} \frac{x^n}{n^2},
$$ converges to some $f(x)$, and that $f(x)$ is continuous, differentiable, and integrable on $[-1,1]$.
I know how to show that $f(x)$ is continuous, since each $f_n(x)$ is continuous, and $f_n(x)$ converges uniformly. Because each $f_n(x)$ is also integrable, I can also show $f(x)$ is integrable.
The trouble I'm having is proving that $f(x)$ is differentiable. I need to show that the series of derivatives converges uniformly. However, I don't think I can use the Weierstrass M-Test in this scenario. Any ideas?
 A: Two theorems we will use about uniform convergence:
Theorem 1 Let $f,f_n:X\to \mathbb{R}$. If $f_n$ is continuous at $a\in X$  for $n=0,1,...$ and $\sum_{k=0}^{\infty}f_k$ converges uniformly then $\sum_{k=0}^{\infty}f_k(x)$ is continuous at $a$
Theorem 2 Let $f_n:[a,b]\to \mathbb{R}$. If $f_n$ is integrable for $n=0,1,...$ and $\sum_{k=0}^{\infty}f_k$ converges uniformly then 
\begin{equation}\int_{a}^{b}\sum_{k=0}^{\infty}f_k=\sum_{k=0}^{\infty}\int_{a}^{b}f_k \end{equation}
Now as the OP said the series
$$\sum_{n=1}^{\infty}\frac{x^n}{n^2}$$
converges uniformly in $[-1,1]$ and to a function $f:[-1,1]\to \mathbb{R}$. The same can be said for the derived series
$$\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$$
which converges unifomly in $(-1,1)$ to some $g$. That $g$ is continuous and so
\begin{equation}\int_{a}^{x}g=\sum_{n=1}^{\infty}\int_{a}^{x}\frac{t^{n-1}}{n}\, dt= \sum_{n=1}^{\infty}\frac{x^n}{n^2}-\frac{a^n}{n^2}=f(x)-\sum_{n=1}^{\infty}\frac{a^n}{n^2}\end{equation}
for $-1<a<x<1$. Therefore, by the 1st fundumental theorem of calculus, $f$ is differentiable in $(-1,1)$ and $f^{\prime}=g$.
Since $f$ is continuous in $[-1,1]$ and differentiable in $(-1,1)$, by the Mean Value Theorem for arbitrary small $h>0$,
$$\exists \xi\in (-1,-1+h):f^{\prime}(\xi)=\frac{f(-1+h)-f(-1)}{h} $$
As $h\to 0^+$, $\xi \to -1^+$ and $f(x)\to \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$ which converges and so $f$ is differentiable at $-1$.
Because $\lim_{x\to 1^-}f^{\prime}(x)=\sum_{n=1}^{\infty}\frac{1}{n}=+\infty$, $f$ is not differentiable at $1$. Remember that $f^{\prime}$ is only allowed to have essential discontinuities and not poles.
Therefore, $f$ is differentiable only in $[-1,1)$ and
$$f^{\prime}(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}$$
