Derivative of $ f(x) = \sum\limits_{n=1}^{\infty} \frac{1}{n^2}e^{-n^2x^2}$ at 0 Let $ f(x) = \sum_{n=1}^{\infty} \frac{1}{n^2}e^{-n^2x^2}$
I know that the serie of the derivatives $g(x) = \sum_{n=1}^{\infty} -2xe^{-n^2x^2}$ is uniformely convergent on every set $]-\infty;-b] \cup [b;+\infty[ , b>0 $.
This implies that $f$ is differentiable on $\mathbb{R} \setminus \{0\}$.
My question is, is it differentiable on $0$ ?
I tried to apply L'hopistal rule but I cannot figure out if $g(x)$ has a limit when $x \to 0^+$ or $x \to 0^{-}$.
I also tried to compute the derivative at $0$ but I don't see how to carry the computations.
Many thx for any help
 A: The function$$
f(x)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}e^{-x^{2}n^{2}}
$$
has a maximum at $x=0$, since $e^{-x^{2}n^{2}}\leq e^{0}=1$, so if $f$ is
differentiable at $x=0$, necessarily, $f^{\prime }(0)=0$. For $m\in \mathbb{N
}$ we have$$
\frac{f(0)-f(\frac{1}{m})}{\frac{1}{m}-0}=m\sum_{n=1}^{\infty }%
\frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}})\geq m\sum_{n=1}^{m}
\frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}})
$$
Consider the function $g(t)=1-e^{-t}-\frac{t}{2}$. Note that $g(0)=0$ and $%
g(1)=\frac{1}{2}-e^{-1}>0$. Moreover, $g^{\prime }(t)=e^{-t}-\frac{1}{2}>0$
for $t<\log 2$. Thus, $g>0$ for $t\in (0,1)$ and so $1-e^{-t}\geq \frac{t}{2}
$ for $t\in \lbrack 0,1]$. Using this inequality with $t=\frac{n^{2}}{m^{2}}
\leq 1$ we get that$$
m\sum_{n=1}^{m}\frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}})\geq 
m\sum_{n=1}^{m}\frac{1}{n^{2}}\frac{1}{2}\frac{n^{2}}{m^{2}}=\frac{
1}{2}.
$$
This shows that$$
\frac{f(0)-f(\frac{1}{m})}{\frac{1}{m}-0}\geq \frac{1}{2},
$$
that is $\frac{f(\frac{1}{m})-f(0)}{\frac{1}{m}-0}\leq -\frac{1}{2}$. Hence, 
$$\liminf_{x\rightarrow 0^{+}}\frac{f(x)-f(0)}{x-0}\leq -\frac{1}{2},
$$
which implies that $f$ cannot be differentiable at $x=0$.
