Prove $\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}\arctan(\frac{x}{\sqrt{n}})$ converges and defines a continuously differentiable function

Question

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?

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 $x$.