# 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. – David Mitra Nov 2 '15 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. – Tenno Nov 2 '15 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.