Prove $\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\arctan\big(\frac{x}{\sqrt{n}}\big)$ converges for all $x,$ and defines a continuously differentiable function on $\mathbb{R}.$

By Leibniz test, the series converges for all $x.$

For $x=0,$ we get series of $0$'s, and the series converges.

Let us show the series of the derivatives converges uniformly:

$$\bigg(\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\arctan\bigg(\frac{x}{\sqrt{n}}\bigg)\bigg)'=\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}\sqrt{n}\big(1+\frac{x^2}{n}\big)}=\sum_{n=1}^\infty \frac{(-1)^n}{x^2+n}$$

$\sum(-1)^n$ is bounded uniformly and $\frac{1}{x^2+n}$ decreasing to $0,$ so by Dirichlet's test the series of derivatives converges uniformly.

By term-by-term differention theorem we conclude the original series is continuously differentiable.

Is that correct?


I wasn't aware there was a Dirichlet test for series of functions. After reading the criterion here, it seems that you forgot to check that $\frac{1}{x^2+n}$ decreases uniformly to $0$ (easy).

Anyway, $\sum_{n=1}^\infty \frac{(-1)^n}{x^2+n}$ can be proved to be converge uniformly by a different argument: let $N\geq 0$ and $x\in \mathbb R$. $$\left| \sum_{n=N+1}^\infty \frac{(-1)^n}{x^2+n} \right| \leq \frac{1}{x^2+N+1} $$ by a classic result on alternating series (of real numbers). And $ \frac{1}{x^2+N+1}\leq \frac{1}{N+1}$ which goes to $0$ as $N\to \infty$ and does not depend on $x$.

  • $\begingroup$ Unrelated but caught interest: >(of real numbers) Is it also true for complex number (adding absolute value to r.h.s)? $\endgroup$ – Rab Aug 18 '17 at 14:43
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    $\begingroup$ @RabMakh The usual context is $\sum (-1)^n a_n$ with $a_n$ decreasing to $0$. If $a_n\in \mathbb C$, how do you define "decreasing" ? I guess you can require $|a_n|$ decreasing to $0$. But for example, with $a_n=\frac{(-1)^n}n$, the series $\sum (-1)^n a_n$ diverges, though $|a_n|$ decreases to $0$. $\endgroup$ – Gabriel Romon Aug 18 '17 at 14:50
  • $\begingroup$ I didn't understand your argument, could you please elaborate why $\frac{1}{N+1} \rightarrow 0$ implies uniform continuity? Other then that, is my proof correct? $\endgroup$ – Itay4 Aug 18 '17 at 14:53
  • $\begingroup$ @Itay4 if $\forall x, |f(x)-f_n(x)|\leq a_n$ and $a_n \to 0$, then $f_n$ converges uniformly to $f$. $\endgroup$ – Gabriel Romon Aug 18 '17 at 14:55
  • $\begingroup$ I see, thanks. Other then that, is my proof correct? $\endgroup$ – Itay4 Aug 18 '17 at 14:58

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