$\sum_{n=1}^{\infty} -2xe^{-n^2x^2}$

Question

How could I prove that the following serie is not uniformely convergent in [-1;1] :

$f(x) = \sum_{n=1}^{\infty} -2xe^{-n^2x^2}$

Edit : I found the answer,

if it is uniformely convergent then we can find a sequence $N_1, N_2 ,...,N_k,...$ such that for any sequence $(x_{k}) \in [-1;1], |\sum_1^{N_k} f_n(x_{k}) - \sum_1^{N_k -1} f_n(x_{k}) | \leq \frac{1}{k^2}$

But if we consider the sequence $x_k=\frac{1}{N_k}$ then $|\sum_1^{N_k} f_n(x_{k}) - \sum_1^{N_k -1} f_n(x_{k}) | = \frac{2}{N_k}$.

Edit 2 : The answer i found is wrong i am still stuck

Answer 1

You have to find some $\epsilon>0$ such that for any $N>0$ there exists an $x_N$ such that $|\sum_{n=N+1}^{\infty}-2x_N\exp(-n^2x_N^2)|>\epsilon$.

To prove this, let $x_N:=1/N$. Then

$$ |\sum_{n=N+1}^{\infty}-2x_N\exp(-n^2x_N^2)|=\sum_{n=N+1}^{\infty}2/N\exp(-n^2/N^2)\\ \geq \sum_{n=N+1}^{2N}2/N\exp(-4)=2\exp(-4). $$ Hence, you may conclude using $\epsilon:=2\exp(-4)$.