Point of differentiability There is a function defined by a functional row $$\sum_{n=1}^{+∞}\frac{e^{-n^{2}x^{2}}}{n^{2}}$$
Need to prove that this function is not differentiable at zero.
 A: I think we need a more explicit bound than the one appearing in the other answer, since if we replace $e^{-n^2 x^2}$ with $e^{-\sqrt{n}x^2}$ or $e^{-nx^2}$ we do have a differentiable function at the origin. The point is that the formal derivative of our function is $-2x\sum_{n\geq 0}e^{-n^2 x^2}$, and $\sum_{n\geq 0}e^{-n^2 x^2}$ is unbounded as $x\to 0$. On the other hand the (lack of) differentiability at the origin of $f(x)$ depends on how fast the series $\sum_{n\geq 1}e^{-n^2 x^2}$ converges to $+\infty$ as $x\to 0^+$. With more rigor: we may safely assume that $x$ belongs to $\mathbb{R}^+$ since our function is even. By setting
$$ f(x)=\sum_{n\geq 1}\frac{e^{-n^2 x^2}}{n^2} $$
we have that $f(x)$ is a continuous function, since it is a totally convergent series of continuous functions. We have $f(0)=\zeta(2)=\frac{\pi^2}{6}$ and for any $h>0$
$$ \frac{f(h)-f(0)}{h}=-\sum_{n\geq 1}\frac{1}{n^2}\cdot\frac{1-e^{-n^2 h^2}}{h}$$
is negative. On the other hand for any $x\in(0,1]$ we have $x^2>1-e^{-x^2}>\frac{x^2}{2}$ and for any $x>1$ we have $(1-e^{-1})<1-e^{-x^2}<1$, so 
$$ \sum_{n=1}^{N}\frac{1}{n^2}\cdot\frac{1-e^{-n^2 h^2}}{h}=\sum_{n\leq 1/h}\frac{1}{n^2}\cdot\frac{1-e^{-n^2 h^2}}{h}+\sum_{1/h<n\leq N}\frac{1}{n^2}\cdot\frac{1-e^{-n^2 h^2}}{h}$$
is greater than
$$ \sum_{n\leq 1/h}\frac{n^2 h^2}{2n^2 h}+(1-e^{-1})\sum_{1/h<n<N}\frac{1}{hn^2}\approx \frac{1}{2}+(1-e^{-1})\left(h-\frac{1}{N}\right). $$
In particular in any right neighbourhood of the origin there is a point in which $f'(x)$ is less than a negative constant. This contradicts the differentiability of $f$ at the the origin: since $f$ is an even function, its differentiability at the origin would automatically imply $f'(0)=0$, and derivatives have the mean value / Darboux property.

A more advanced approach: by the Poisson summation formula and the reflection formula for the Jacobi $\Theta$ function, in a neighbourhood of the origin
$$ \sum_{n\geq 1} e^{-n^2 x^2} \approx \frac{1}{2}\sqrt{\frac{\pi}{1-e^{-x^2}}}\approx \frac{\sqrt{\pi}}{2|x|} $$
hence
$$ \lim_{x\to 0^\pm}\frac{d}{dx}\sum_{n\geq 1}\frac{e^{-n^2 x^2}}{n^2}=\color{red}{\mp\sqrt{\pi}}$$
and $f(x)$ essentially behaves like $\zeta(2)\,e^{-\frac{6|x|}{\pi\sqrt{\pi}}-\frac{12x^2}{\pi^3}}$ as $x\to 0$.
