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

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.

## 1 Answer

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.

• I tried this way, but I wasn’t sure that this is possible. Because I tried to demostrate tha the series was a Cauchy series, but also splitting the series in those two, it was useless. Therfore is it always possible to use that method when i have a series like that? – Luca Sansilvestri May 28 '20 at 9:03
• If two series are uniformly convergent the their sum is also uniformly convergent. This is always true. @LucaSansilvestri – Kavi Rama Murthy May 28 '20 at 9:05
• Ok, i did not realize that it was that simple. thanks – Luca Sansilvestri May 28 '20 at 9:07