uniform convergence of functional series $\sum_{k=1}^{\infty} {(-1)^{k} \frac{k+\sin(x)}{k^{2}}}$

Question

I tried to solve the exercise that ask to define the convergence and the uniform convergence of the functional series $$\sum_{k=1}^{\infty} {(-1)^{k} \frac{k+\sin(x)}{k^{2}}}$$ it is easy to prove that the series converge on $\mathbb{R}$ using Leibniz' rule, but I don’t know how to prove that the series converge uniformly, because the series doesn’t converge totally.

Answer 1

Split this into two series. $ \sum\limits_{k=1}^{\infty}\frac {(-1)^{k}} k$ is convergent by Alternating Series Test. So then given series converges uniformly iff $ \sum\limits_{k=1}^{\infty} \frac {(-1)^{k} \sin x} {k^{2}}$ is uniformly convergent. This is indeed true my M-test since $|\sin x| \leq 1$ and $\sum \frac 1 {k^{2}}$ is convergent.