Function series and term-by-term derivative I have the series of functions $(x \in \mathbb{R}, p \in \mathbb{R}^+)$:
$$f(x) = \sum_{n=1}^\infty \frac{\log(1+ n^2x^2)}{n^p}.$$
I want to determine values of $p$ such that (1) series converges and (2) series can be differentiated term-by-term for all $x \in \mathbb{R}$.
$$f'(x) = \sum_{n=1}^\infty \frac{2n^2x}{n^p(1 + n^2x^2)}.$$
For (1) I know that $$\frac{\log(1 + n^2x^2)}{n^p} < \frac{n^2x^2}{n^p} = \frac{x^2}{n^{p-2}}$$
so the series converges if $p-2 > 1$ and $p > 3$, but I think it may converge for values less than $3$.
For (2) I think that the second series is uniformly convergent on any closed interval for $p > 1$ so term-by-term differentiation works.
Is this correct?
 A: First, the series for $f(x)$ converges for all $x \in \mathbb{R}$ when $p > 1$. 
Let $p = 1 + \epsilon$ where $\epsilon > 0$.  Since the series $\sum \frac{1}{n^{1+ \epsilon/2}}$ converges and
$$\lim_{n \to \infty} n^{1 + \epsilon/2}\frac{\log(1 + n^2x^2)}{n^p} =  \lim_{n \to \infty} \frac{\log(1 + n^2x^2)}{n^{\epsilon/2}} = 0$$
we have convergence by the limit comparison test.
For any compact interval $[a,b]$ that excludes $0$ we have
$$\left|\frac{2n^2x}{n^p(1 + n^2x^2)}\right| \leqslant  \frac{2n^2|x|}{n^{p+2} x^2}\leqslant \frac{2}{n^p \min(|a|,|b|)},$$
and the series of derivatives converges uniformly if $p >1$. Thus for all $p > 1$ and all $x \neq 0$ we have
$$f'(x) = g(x) = \sum_{n=1}^\infty \frac{2n^2x}{n^p(1 + n^2x^2)} .$$
It remains to consider when $f'(0) = g(0) = 0.$ On any interval $[a,b]$ with $0 \in [a,b]$ we have
$$\left| \frac{2n^2x}{n^p(1 + n^2x^2)} \right| = \frac{1}{n^{p-1}}\frac{2n|x|}{1 + n^2x^2} \leqslant \frac{1}{n^{p-1}},$$
and the series converges uniformly for $p > 2$ $(p-1 > 1)$.
Thus, the series can be differentiated termwise for all $x$ if $p > 2$ and $f'(0) = g(0) = 0.$  I suspect $f'(0) \neq g(0)$ if $1 < p \leqslant 2$.
