Show that $\sum\limits^{\infty}_{n=1} (-1)^n \frac {x^2+n} {n^2}$, $x \in \mathbb R$, converges uniformly on every bounded interval Show that  $\sum\limits^{\infty}_{n=1} (-1)^n \frac {x^2+n} {n^2}, x \in \mathbb R$ converge uniformly on every bounded interval $I \subseteq \mathbb R$.


I've already shown that the series does not converge absolutely for $x \in \mathbb R$. I guess, I proved this to conclude that the series doesn't have a majorant series which would imply that the series converged uniformly (viewed as a series of functions) and absolutely as a tradional series for $x \in \mathbb R$.

I know I must show that the $s_k = \sum^{k}_{n=1} (-1)^n \frac {x^2+n} {n^2}$ converge uniformly, but I haven't had any luck in showing that.
Well, the simpler the better...
$$
\sum_{n\geqslant N}(-1)^n \frac {x^2+n} {n^2}
=
x^2\cdot\sum_{n\geqslant N}\frac{(-1)^n}{n^2}
+
\sum_{n\geqslant N}\frac{(-1)^n}n
$$
Edit: Both series in the RHS are alternating series hence the absolute value of a given rest is bounded by the absolute value of the first term in the rest, that is,
$$
\left|\sum_{n\geqslant N}\frac{(-1)^n}{n^2}\right|\leqslant\frac1{N^2},\qquad
\left|\sum_{n\geqslant N}\frac{(-1)^n}n\right|\leqslant\frac1N.
$$
We have $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}=-\ln2$$

$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$$

Hence $$\sum_{n=1}^{\infty} (-1)^n\frac{x^2+n}{n^2}=-\frac{\pi^2x^2}{12}-\ln2$$