# Uniforme convergence series : $x\in(0,1)\\ \sum_{n\ge 0} \frac{xe^{n^2}}{1+n^2x^2e^{n^2}}$

how prove Uniforme convergence series :

$$x\in(0,1)\\ \sum_{n\ge 0} \frac{xe^{n^2}}{1+n^2x^2e^{n^2}}$$

Let :$$f_n(x)= \frac{xe^{n^2}}{1+n^2x^2e^{n^2}}$$ $$a_n=\text{sup}_{x\in [0,1]}{|f_n(x)|}$$

we have : $$f'_n(x)=\frac{e^{n^2}(1-n^2x^2e^{n^2})}{(1+n^2x^2e^{n^2})^2}$$

if $$x \ge \frac{1}{ne^{n^2/2}} :f_n(x) \; \; \text{deacreasing}$$

if $$x \le \frac{1}{ne^{n^2/2}} :f_n(x) \; \text{increasing}$$

therfore : $$a_n = f_n(\frac{1}{ne^{n^2/2}})=\frac{e^{n^2/2}}{n(1+n)}$$

and we have series :$$\sum_{n\ge 0} \frac{e^{n^2/2}}{n(1+n)}$$ diverge so : series $$\sum_{n\ge 0} \frac{xe^{n^2}}{1+n^2x^2e^{n^2}}$$ not Uniforme convergence

is my methode correct ?? or not

Your method is not correct : you only proved that the series is not normally convergent over $$(0,1)$$, but it is not sufficient to prove that it is not uniformly convergent. However, you got
$$||f_n||_{\infty} = f_n \left(\dfrac{1}{ne^{n^2/2}} \right) = \dfrac{e^{n^2/2}}{2n}$$
which leads to $$\lim_{n \rightarrow +\infty} ||f_n||_\infty = +\infty$$.
Hence the sequence $$(f_n)_{n \in \mathbb{N}}$$ does not converge uniformly to $$0$$ over $$(0,1)$$, and hence, the series $$\sum f_n$$ does not converge uniformly over $$(0,1)$$.