Why $f(x)=\Sigma_{n=0}^{\infty}\frac{x^n}{n\ln(n)}$ $\in c^n$

Question

consider $f(x)=\sum_{n=0}^{\infty}\frac{x^n}{n\ln(n)}$ and $h(x)=\sum_{n=0}^{\infty}\frac{\sin(nx)}{1+n^4}$.

why $f(x)\in c^n$ in $[\frac{-1}{8},\frac{1}{9}]$ and $h(x)\in c^2$ in $\mathbb{R}$

I think h(x) is a periodic function and I think it can get derivative infinte times.

I just not sure I understand how to show that a function of series can get derivative specific times.

Answer 1

hint

$f(x)$ is the sum of a power series whose radius of convergence is $$\lim_{n\to+\infty}\frac{(n+1)\ln(n+1)}{n\ln(n)}=1$$ thus $ f $ is $C^\infty $ at $(-1,1)$ and therefore at $ [-\frac 18 \frac 19] $.

$h $ is the sum of a series of functions satisfying

$$(\forall x\in \Bbb R)\;\;|\frac{\sin(nx)}{2+n^4}|\le \frac{1}{n^4}$$ thus the series is uniformly convergent at $\Bbb R$. The series of the derivatives satisfies $$(\forall x\in\Bbb R)\;\; |\frac{n\cos(nx)}{1+n^4}|\le \frac{1}{n^3}$$

thus, $h$ is differentiable at $\Bbb R$ and $$h'(x)=\sum_{n=0}^\infty \frac{n\cos(nx)}{1+n^4}$$ but

$$|\frac{-n^2\sin(nx)}{1+n^4}|\le \frac{1}{n^2}$$

thus $h'$ is differentiable at $\Bbb R$ and $$h''(x)=\sum_{n=0}^\infty \frac{-n^2\sin(nx)}{1+n^4}$$

so, $ h''$ is continuous at $ \Bbb R$.