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?


1 Answer 1


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, 2017 at 14:43
  • 1
    $\begingroup$ @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 $0$. $\endgroup$ Aug 18, 2017 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, 2017 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 $f$. $\endgroup$ Aug 18, 2017 at 14:55
  • $\begingroup$ I see, thanks. Other then that, is my proof correct? $\endgroup$
    – Itay4
    Aug 18, 2017 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.