How to show that $\sum_{n=1}^\infty (-1)^n\frac {x^2+n}{n^2}$ is uniformly convergent?

Question

Show that $\sum_{n=1}^\infty (-1)^n\frac {x^2+n}{n^2}$ is uniformly convergent on arbitrary interval.

I wanted to use M test for arbitrary [a,b] $|(-1)^n\frac {x^2+n}{n^2}|\le|(\frac{b^2+n}{n^2})|\le (\frac{b^2+n^2}{n^2})|=M_n$

it seems $\sum M_n=b^2\sum\frac{1}{n^2}+1=\frac {b^2\pi^2}{6}+1$ is it correct?

Answer 1

Dirichlet's test shows that the series is uniformly convergent on a bounded interval. The term $(-1)^n$ has bounded partial sums and the term $(n+x^2)/n^2$ converges monotonically and uniformly to $0$ for $x \in [a,b]$.