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

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.

• The $M$-test won't apply; but see this. Nov 2, 2015 at 8:46
• Am I misunderstanding something, or can you not use the Alternating Series Test to demonstrate the convergence of "B"? Though this probably won't help, since the suggested inequality isn't true. Nov 2, 2015 at 9:23

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.