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

Question

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

Answer 1

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 $(0,1)$.