Explore convergence of the series on $E = (0, 1)$: $$\sum_{k=3}^{\infty} \frac{1}{2^k \tan(\pi x/k)} $$
I had an idea to use Dirichlet's test as $\frac{1}{2^k} \to 0$ and so I need to prove that $\displaystyle \forall N :\left|\sum_{k=3}^{N} \frac{1}{\tan(\pi x/k)} \right| \leq M$ but couldn't do that. So maybe I'm going in wrong direction. I would appreciate any help.
$|\tan(z)| \ge |z/2|$ near $z = 0$. This is because the derivative of $\frac d {dz} \tan(z) |_{z = 0} = 1$.
So for large enough $k$, $$|\frac{1}{\tan(\pi x /k)}| \le |\frac{2 k}{\pi x}|$$ $$|\frac{1}{2^k \tan(\pi x /k)}| \le |\frac{k}{\pi x 2^{k-1}}| $$
So the sum is absolutely convergent for all $x$ since $\frac{k}{2^{k-1}}$ is summable in $k$.
I think a similar bound that $|\tan(z)| \le \frac 3 4 z$ can show that the convergence is not uniform. For large enough $k$,
$$|\frac{1}{\tan(\pi x /k)}| \ge |\frac{4 k}{3\pi x}|$$ $$|\frac{1}{2^k \tan(\pi x /k)}| \ge \|\frac 1 x\frac{4k}{3\pi 2^{k-1}}| $$
So the tail of the sum, as a function of $x$, can be made as large as we want by choosing $x$ close enough to $0$.