I have this series $\sum_{n=1}^{+\infty} (-1)^n \frac{x^2+n}{n^2}$.
The series pointwise converges for Leibniz in all R but doesn't converge absoluty because $\sum_{n=1}^{+\infty}|\frac{x^2+n}{n^2}|$ diverges ($\frac{x^2+n}{n^2} \sim {1 \over n}$ and $\sum_{n=1}^{+\infty}{1 \over n}$ diverges).
For uniformly convergence $\sum_{n=1}^{+\infty}\sup|\frac{x^2+n}{n^2}|=+\infty$ so there isn't uniformly convergence in all R. But in intervals as $[-a,a],a>0$?
For each natural $N$,$$\left|\sum_{n=N}^\infty(-1)^n\frac{x^2+n}{n^2}\right|\leqslant\frac{x^2+N}{N^2}.$$If $x\in[-a,a]$, this is smaller than or equal to $\frac{a^2+N}{N^2}$ and $\lim_N\frac{a^2+N}{N^2}=0$.