Prove that $\sum\limits_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2}$ converges uniformly, but not absolutely

Question

My Work: i) Uniform Convergence (By Weierstrass M-Test): I am attempting to show that the series converges uniformly on the interval $I=[-a,a]$, in which:$$\sum_{n=1}^\infty(-1)^n\frac{x^2+n}{n^2}\le\sum_{n=1}^\infty(-1)^n\frac{a^2+n}{n^2}=\varepsilon_0$$ Suppose that $\sum\limits_{n=1}^\infty\frac{x^2+n}{n^2}$ will be called "A" and $\varepsilon_0$ will be called "B". By the M-test, if B converges, then A converges uniformly on the interval $I$ defined above. I am having difficulty proving that B converges, however. I have tried both the root and ratio tests, and they have been unhelpful. For this segment, could someone confirm my logic / provide a hint towards the convergence of B? ii) Absolute Convergence: I believe that this function does not converge absolutely because$\sum\limits_{n=1}^\infty\frac{x^2+n}{n^2}$ is a divergent sum as n approaches infinity. Bit of a trivial solution here but I believe it's sufficient.

Answer 1

Your argument is correct for the non-absolute convergence. Indeed, the series $\sum_{n\geqslant 1}x^2/n^2$ is convergent while $\sum_{n\geqslant 1}1/n$ is divergent, hence $\sum_{n\geqslant 1}\left(x^2/n^2+1/n\right)$ is divergent.

For the uniform convergence, define $s_n(x):=\sum_{j=1}^n(-1)^j(x^2+j)/j^2$. Defining $S_n(x):=\sum_{j=1}^n(-1)^j\frac{x^2}{j^2}$, we have the equality $$s_n(x)=S_n(x)+\sum_{j=1}^n(-1)^j/j.$$ Since the series $\sum_{j=1}^{+\infty}(-1)^j/j$ is convergent, it suffices to establish the uniform convergence of the sequence $\left(S_n\right)_{n\geqslant 1}$ on $[-a,a]$, which can be done by the Weierstrass $M$-test.