Proving uniform convergence of this series I'm trying to prove that the following series
$$\sum_{n=1}^{\infty}(-1)^n \dfrac{x^2 + n}{n^2}$$
converges uniformly on every finite interval $I$ in $\mathbb{R}$.
In the previous exercice I've shown that the series is not absolute convergent for any $x \in \mathbb{R}$, so this might be of use, but I'm not sure how.
I've attempted to use the Weierstrass M-test and tried to find some $M_n \geq \dfrac{x^2 + n}{n^2} \; \forall x \in I$ such that $\sum M_n$ converges. But I realised that this would imply that $M_n \geq \frac{1}{n}$ and thus $\sum M_n$ cannot converge.
I also tried finding the partial sums for the series, but this also turned out to be pretty tough.
How would I go about showing uniform convergence on $I$?
 A: 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) 
A: Let $M>0$ such that $I \subset [-M,M]$
Let $$\begin{array}{ccccc}
f & : & I & \to & \mathbb R \\
 & & x & \mapsto & \sum_{n=1}^{\infty}(-1)^n \dfrac{x^2 + n}{n^2} \\
\end{array}$$
For $m\in \mathbb N$, let  $$\begin{array}{ccccc}
f_m & : & I & \to & \mathbb R \\
 & & x & \mapsto & \sum_{n=1}^{m}(-1)^n \dfrac{x^2 + n}{n^2} \\
\end{array}$$
Now for $x \in I$, because the series is alternate,
$$|f(x)-f_m(x)|\leq \dfrac{x^2 + n+1}{(n+1)^2} \leq \dfrac{M^2 + n+1}{(n+1)^2}$$
Because $\frac{M^2 + n+1}{(n+1)^2} \to 0$ and is independent on $x$, uniform convergence is granted.
