The problem is
Prove that the series $\sum_{n=1}^\infty (-1)^n\frac{x^2+n}{n^2}$ converges uniformly in every bounded interval, but does not converge absolutely for any value of $x$.
My attempt is:
(a) Let $[a,b]\subset\mathbb{R}$. Let $\epsilon>0$. Choose $N\in\mathbb{N}\ni \forall n\geq N$, $\sum_{n=k}^\infty\frac{1}{k^2}<\frac{\epsilon}{2b^2} \qquad\text{and}\qquad \sum_{k=n}^\infty\frac{(-1)^k}{k}<\frac{\epsilon}{2}.$ Let $n\geq N$. Then $\sum_{k=n}^\infty (-1)^k\frac{x^2+k}{k^2}=\sum_{k=n}^\infty\left((-1)^k\frac{x^2}{k^2}+(-1)^k\frac{1}{k^2}\right)\leq\sum_{k=n}^\infty(-1)^k\frac{x^2}{k^2}+\sum_{k=n}^\infty(-1)^k\frac{1}{k^2}$ $\leq\sum_{k=n}^\infty \frac{b^2}{k^2}+\sum_{k=n}^\infty(-1)^k\frac{1}{k^2}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ $\therefore \sum_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2} \quad\text{converges uniformly on }[a,b]$
(b) $\sum_{n=1}^\infty\left|(-1)^n\frac{x^2+n}{n^2}\right|=\sum_{n=1}^\infty \left|\frac{x^2+n}{n^2}\right|\leq\sum_{n=1}^\infty\frac{1}{n}$ hence the series is not absolutely convergent.
Is my procedure correct?
Thanks for your help