Nonexistence of derivative of series of functions From this question Function series and term-by-term derivative it's shown the function $(x \in \mathbb{R}, p \in \mathbb{R}^+)$:
$$f(x) = \sum_{n=1}^\infty \frac{\log(1+ n^2x^2)}{n^p}$$
can be differentiated term-by-term for all $x \in \mathbb{R}$ if $p > 2$ and for all $x \neq 0$ when $1 < p  \leqslant 2:$
$$f'(x) = \sum_{n=1}^\infty \frac{2n^2x}{n^p(1 + n^2x^2)}.$$
But it is claimed $f'(0)$ either does not exist or is not obtained from the term-by-term derivative when $1 < p \leq 2$.
Is it true and how can this be shown?
 A: The derivative $f'(0)$ does not exist if upper and lower Dini derivatives are not all equal -- for example, if we have
$$D^+f(0)= \limsup_{h \to 0+}\frac{f(h) - f(0)}{h} \neq \liminf_{h \to 0-}\frac{f(h) - f(0)}{h}= D_{-}f(0). $$
The function in question is even, $f(x) = f(-x)$ and $f(0) = 0$. Hence,
$$\liminf_{h \to 0-}\frac{f(h) - f(0)}{h}  = \liminf_{h \to 0+}\frac{f(-h)}{-h} = \liminf_{h \to 0+}\left(-\frac{f(h)}{h}\right) = - \limsup_{h \to 0+} \frac{f(h)}{h},$$
and $D^+f(0) = - D_{-}f(0),$ whence, $f'(0)$ does not exist unless $D_+f(0) = 0$.
However,
$$\frac{f(h)}{h} = \sum_{n=1}^\infty \frac{\log(1 + n^2h^2)}{n^ph} \geqslant \sum_{n=m}^\infty \frac{\log(1 + n^2h^2)}{n^ph} $$ 
Choosing a sequence $h_m = m^{-1}$ we have 
$$\sum_{k=m}^\infty \frac{\log(1 + n^2m^{-2})}{n^pm^{-1}}  \geqslant m \log 2\sum_{n=m}^\infty\frac{1}{n^p}  \geqslant m \log 2\int_{m}^\infty\frac{dx}{x^p} = \frac{m^{2-p}\log 2}{p-1}, $$
and if $1 < p \leqslant 2$,
$$\limsup_{h \to 0+} \frac{f(h)}{h} \geqslant \lim_{m \to \infty}\frac{m^{2-p}\log 2}{p-1} \neq 0.$$
Therefore, $f'(0)$ does not exist.
