Showing differentiability of a power series

Question

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 $[-a,a]$.

Not if $a=1$. It diverges at $x=1$. – Robert Israel May 22 '20 at 02:07 — Robert Israel, May 22 '20 at 02:07
My mistake - I've fixed it now, thanks. – Matt May 22 '20 at 02:29 — Matt, May 22 '20 at 02:29

score 1 · Answer 1 · answered May 22 '20 at 01:03

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 $[-a,a]$.

Thank you! This helped me greatly. – Matt May 22 '20 at 02:30 — Matt, May 22 '20 at 02:30

1 Answers1