I am given a function expressed as
$h(x) = \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} xe^{-n^4x^2}$
I want to express the derivative $h'(x), x \in (0,\infty)$ as a series. From what I understand I need to show three things in order to $(\sum_{n=1}^{\infty} f_n(x))' = \sum_{n=1}^{\infty} f'_n(x)$ being true. and those are:
$f'_n(x) \in \mathbb{R}, \forall n \in \mathbb{N}$: Which should be as simple as computing the derivative $f'_n(x) = e^{-n^4x^2}(1-2n^4x^2)$ and seeing there is no problem there.
$\exists x_0 \in (0,\infty)$ such that $\sum_{n=1}^{\infty} f_n(x)$ converges: which is again simple enough, we can pick i.e. $x_0 = 1$, then the series $\sum_{n=1}^{\infty} e^{-n^4}$ surely converges.
$\sum_{n=1}^{\infty} f'_n(x)$ converges uniformly for $ x\in (0,\infty)$: and here is my problem. I know that if $\sum_{n=1}^{\infty} sup_{x\in(0,\infty)}|f'_n(x)| = \sum_{n=1}^{\infty} sup_{x\in(0,\infty)}|e^{-n^4x^2}(1-2n^4x^2)|$ converges then $\sum_{n=1}^{\infty} f'_n(x)$ converges. To obtain the supremum I computed $f''_n(x)=0$ and got $ x\in\{{0,\sqrt{\frac{3}{2n^4}}}\} $ and the $lim_{x \rightarrow \infty} f_n(x) = 0$. So the supremum is for $x=0$. Fianlly $\sum_{n=1}^{\infty} sup_{x\in(0,\infty)}|f'_n(x)| = \sum_{n=1}^{\infty} |f'_n(0)| = \sum_{n=1}^{\infty} 1 $ and this series diverges.
So in the end this doesn't give me any information. What should I do? Am I doing this right? Is the locally uniform convergence good enough in this instance? I believe for $x\in(\epsilon, \infty)$ for$\space small \space \epsilon \in \mathbb{R} \space $ it would converge, right? If it doesn't converge and the derivative doesn't exist how do I show that?
Thank you.
