I'm struggling to show differentiability of a sumfunction $f:[-a,a]\rightarrow\mathbb{R}$. I'm dealing with the following series.
$$\sum_{n=1}^\infty \frac{1}{2n}x^{2n} \hspace{25pt}x\in\mathbb{R}$$
and let $0\leq a<1$.
I've already shown that the series is uniform convergent on $[-a,a]$.
Let $$f_n(x)=\frac{1}{2n}x^{2n}$$
for $x>0$,
$$\lim_{n\to+\infty}|\frac{f_{n+1}(x)}{f_n(x)}|=x^2$$
If $|x|<1$, the series converges. If $|x|\ge 1$, it diverges.
The convergence is pointwise at $(-1,1)$.
Let $a\in(0,1)$.
For $ n$ great enough, $ f_n$ is defferentiable at $[-a,a]$ and
$$|f'_n(x)|=|x|^{2n-1}\le a^{2n-1}$$
the series $\sum a^{2n-1}$ is convergent, thus the series $\sum f'_n(x)$ is uniformly convergent at $[-a,a]$.
The sum of the series $\sum f_n(x)$ is then diffetentiable at $[-a,a]$.