First given $x \in \mathbb{R}$:
$$\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2 + n}{n^2}=\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}$$
So if you prove $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent you are ready (since the second sum converges and is constant and not depends on $x$).
$x^2$ is bounded for every finite interval on $I$ by $M_I$:
$$\left|(-1)^n\frac{x^2}{n^2}\right|\leq \frac{M_I}{n^2} $$
So $\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2}{n^2}$ is absolutly convergent.
Uniformity: pointwise $f_m(x)$ converges to:
$$f(x)=x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}$$
So:
$$|f(x)-f_m(x)|=$$
$$\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}+\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq\left|x^2 \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}-x^2 \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq|x^2|\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
$$\leq M_I\left| \sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n^2}- \sum_{n=1}^{m}(-1)^n \dfrac{1}{n^2}\right|+\left|\sum_{n=1}^{\infty}(-1)^n \dfrac{1}{n}-\sum_{n=1}^{m}(-1)^n \dfrac{1}{n}\right|$$
At that point you can complete the demostration (because the convergence is not given by the point $x$ you take)
Let $K \subset (B(0,R) \cap \mathbb{C} \setminus \mathbb{Z})$ be a compact set and set $d(K,\mathbb{Z}) = \delta > 0$.
If $|n| > R$, then for all $x \in K$ we have
$$ |n^2 - x^2| \geq ||n|^2 - |x|^2| = |n|^2 - |x|^2 \geq |n|^2 - R^2. $$
If $|n| \leq R$ then for all $x \in K$ we have
$$ |n^2 - x^2| = |x - n||x - (-n)| \geq \delta^2. $$
Hence,
$$ \sup_{x \in K} \left| \frac{1}{n^2 - x^2} \right| \leq
\begin{cases} \frac{1}{\delta^2} & |n| \leq R, \\
\frac{1}{|n|^2 - R^2} & |n| > R
\end{cases} $$
and so the series converges uniformly on $K$ by the Weierstrass $M$-test. Finish by taking $K = \overline{B(x_0,r)}$ for all $x_0 \in \mathbb{C} \setminus \mathbb{Z}$ and small enough $r$.
Best Answer
If it did, then it would converge uniformly on $(1,\infty)$. But each partial sum $\sum_{n=1}^N\frac1{z^n+1}$ is bounded there, and so, since we are assuming that the convergence is uniform, $\sum_{n=1}^\infty\frac1{z^n+1}$ would be bounded there too, since there is some $N\in\Bbb N$ such that$$(\forall n\in\Bbb N)(\forall z\in(0,\infty)):m\geqslant N\implies\left|\left(\sum_{n=1}^\infty\frac1{z^n+1}\right)-\left(\sum_{n=1}^m\frac1{z^n+1}\right)\right|<1,$$and this in turn, implies that$$(\forall z\in(0,\infty)):\left|\sum_{n=1}^\infty\frac1{z^n+1}\right|<1+\sup_{z\in(1,\infty)}\left|\sum_{n=1}^N\frac1{z^n+1}\right|.$$On the other hand, if $m\in\Bbb N$,$$\lim_{z\to1^+}\sum_{n=1}^m\frac1{z^n+1}=\sum_{n=1}^m\frac12=\frac m2,$$and therefore there is some $w_m\in(1,\infty)$ such that$$\sum_{n=1}^m\frac1{w_m^{\,n}+1}>\frac m4,$$from which it follows that$$\sum_{n=1}^\infty\frac1{w_m^{\,n}+1}>\frac m4.$$But then $\sum_{n=1}^\infty\frac1{z^n+1}$ is unbounded, and so a contradiction was reached.