Let us consider : $f_n : \mathbb{R}\ni x \mapsto \frac{x}{(x^2+n^2)\log(n)}\in\mathbb{R}$ for $n> 1$.
I need to prove that $\sum _{n\ge 0}f_n$ is uniformly convergent. I've already proved that it is not normally convergent by studying the term $(f_n)'$.
Thanks in advance !
We will use Cauchy's test. Let us consider $$ \Big|\sum_{k=n+1}^m f_k(x)\Big|=\sum_{k=n+1}^m \frac{|x|}{(x^2+k^2)\log k} $$ for $m>n$. We can see that $$\begin{eqnarray} \sum_{k=n+1}^m \frac{|x|}{(x^2+k^2)\log k}&\le& \frac{1}{\log n}\sum_{k=n+1}^m \frac{|x|}{x^2+k^2}\\ &\le&\frac{1}{\log n}\int_n^m \frac{|x|}{x^2+y^2}dy\\ &=&\frac{1}{\log n}\int_{\frac{n}{|x|}}^{\frac{m}{|x|}} \frac{1}{1+z^2}dz\\ &\le&\frac{1}{\log n}\int_{0}^{\infty} \frac{1}{1+z^2}dz\\ &=&\frac{\pi}{2\log n} \end{eqnarray}$$ for all $x\in\mathbb{R}$. Thus, we have $$ \lim_{n,m\to\infty}\Big\|\sum_{k=n+1}^m f_k\Big\|_\infty \le\lim_{n\to\infty}\frac{\pi}{2\log n}=0 $$ and Cauchy's test gives the result.