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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Luca Sansilvestri May 28 '20 at 9:03
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    $\begingroup$ If two series are uniformly convergent the their sum is also uniformly convergent. This is always true. @LucaSansilvestri $\endgroup$ – Kavi Rama Murthy May 28 '20 at 9:05
  • $\begingroup$ Ok, i did not realize that it was that simple. thanks $\endgroup$ – Luca Sansilvestri May 28 '20 at 9:07

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