Let $f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n} \arctan\left(\frac{x}{\sqrt n}\right)$. Show that $f(x)$ converges uniformly.
First, it is easy to see that the series converges for every $x$ by Leibniz test.
Now, I'm not so sure how to prove uniform converges. I thought about the fact that $\arctan\left(\frac{x}{\sqrt n}\right)$ is bounded by $\frac{\pi}{2}$, problem is, it's not a supremum but an upper-bound.
I've tried to look for other tests like Weierstrass M-test but it didn't fit here.
EDIT:
I think we should use Cauchy criteria. Let's assume by contradiction it is diverges there is an $\varepsilon > 0$ such that for every $N$ there are $m,n > N$ such that:
$$\left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \arctan(\frac{x}{\sqrt k})\right| \ge \varepsilon$$
Now, for every $x$:
$$\left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \arctan(\frac{x}{\sqrt k})\right| \le \left| \sum_{k=m}^n \frac{(-1)^k}{\sqrt k} \frac{\pi}{2} \right| = \frac{\pi}{2} \left|\sum_{k=m}^n \frac{(-1)^k}{\sqrt k}\right|$$
Since the later series converges by Leibnitz test, it is a Cauchy series and therefore there is an $N$ such that for every $m,n$:
$$\left|\sum_{k=m}^n \frac{(-1)^k}{\sqrt k}\right| < \frac{2\varepsilon}{\pi}$$
And so, we're done.
Could you verify my proof please?
Thanks.