Convergence, continuity and a continuous derivative $\ \sum_{n\geq 1}^{\infty}\frac{nx}{x^2+n^4}$ I have to determine whether this power series (or function):
$$\ f(x)=\sum_{1\leq n}^{\infty}\frac{nx}{x^2+n^4}~~~~x>0$$
is convergent, whether this function is continuous and whether it has a continuous derivative at every point of its interval and the last part seems to be a little tricky for me.
My steps
The uniform convergence of the power series:
$$ \sup_{x\in(0,\infty)} \frac{nx}{x^2+n^4} = \frac{1}{2n} \\ \lim_{x \to 0^+} \frac{nx}{x^2+n^4} = 0$$
It isnt' convergent uniformly.
Almost uniform convergence:
$$ a\in(0, \infty), ~~x\in(a, 5a) \\ \sup_{x\in(a, 5a)} \frac{nx}{x^2+n^4} \geq \frac{5a}{25a^2+n^4}  $$
It isnt' convergent almost uniformly.
Pointwise convergence:
$$ \frac{nx}{x^2+n^4} \leq \frac{nx}{n^4} = \frac{x}{n^3}$$
It is pointwise convergent from comparison to $\sum\frac{1}{n^2}$ and asymptotic similarity.
Since the series is convergent, it represents a continuous function within its radius of convergence.
Question:
What about the derivative at the every point? A differentiable ower series has to be convergent and to have almost uniform or uniform convergetn series of derivatives. Then I should check whether these derivatives are continuous. However, I am stuck at this point:
$$ \lim_{x\to\infty}\frac{n^5-nx^2}{(x^2+n^4)^2}=0 \\ \lim_{x\to 0}\frac{n^5-nx^2}{(x^2+n^4)^2}=0\\ $$
$$(\frac{nx}{x^2+n^4})'= \frac{n^5-nx^2}{(x^2+n^4)^2} \\ x=n^2\\ \sup_{x\in(0,\infty)} \frac{nx}{x^2+n^4} = \frac{1}{2n}\\ $$
It indicates that the power series is not differentiable, from the divergence of $\sum\frac{1}{n}$, therefore it cannot have a continuous derivative. Does that make sense? Or am I wrong in the particuar equation? I would appreciate your hints.
Edits to the post made with Bernard's help.
 A: You make an error in proving continuity:
This is not a power series in the conventional sense; that is, it is not a sum of the form $\sum_{n=0}^{\infty}{a_nx^n}$.  Thus pointwise convergence does not suffice to show that the resulting sum is continuous. (In fact, this is always true; the corresponding result for power series uses that fact that pointwise convergence of a power series implies uniform convergence.)
It can be written as a power series using complex-analytic methods, but your tags suggest you are not familiar with those techniques.
The key to this problem is analyzing the behavior of each term.
For fixed $n$, let $f_n(x)=\frac{nx}{x^2+n^4}$.  Maximizing $f_n$ was a good first step, and you correctly found the maximum to be $\frac{1}{2n}$.  But you ignored the key issue of where the maximum appears.   $f_n$ is maximized at $x=n^2$.
This is the infamous "moving hump": the maximum is relatively large, but increasing $n$ pushes it further and further from the origin.  To be more precise: on any fixed interval $[0,K]$, only finitely many $n$ have $n^2\in[0,K]$, so only finitely many terms of the sum "see" the full maximum.  All others end up "pegged" at $K$, so that their maximum is $\frac{nK}{K^2+n^4}=o(n^{-3})$.
Thus on any $[0,K]$, the convergence is uniform.  This proves continuity.
For the derivative, you should demonstrate convergence in the $C^1$ norm.
We've already shown uniform convergence, so all that remains is to show uniform convergence of $\sum_n{f_n'}$.  I leave it to you to provide the details.
(Hint: Weierstrass $M$-test)
