I want to show the series $$\sum_{n=1}^{\infty}(-1)^n\frac{x^2+n}{n^2}$$
converges uniformly in any bounded interval and here is my justification.
The series $\sum_{n=1}^{\infty}(-1)^n\frac{x^2}{n^2}$ is uniformly convergent in any bounded interval. Let $M=\{|a|,|b|\}$, where $a,b$ are the endpoints of the interval and so we have $$|\frac{(-1)^nx^2}{n^2}|\leq \frac{M^2}{n^2},$$ and so by M-Weierstrass test, we get $\sum_{n=1}^{\infty}(-1)^n\frac{x^2}{n^2}$ is uniformly convergent.
Secondly, $$\sum_{n=1}^{\infty}(-1)^n\frac{n}{n^2}=\sum_{n=1}^{\infty}(-1)^n\frac{1}{n},$$
which is uniformly convergent. Since both sums converge uniformly, we can say teh series
$$\sum_{n=1}^{\infty}(-1)^n\frac{x^2+n}{n^2}$$
converges.
However, the above series is not absolutely convergent, so I'm not sure if rewriting it as a combination of two uniformly convergent series is an accurate argument. I appreciate if you correct me.