Convergence/divergence of $\sum_{n=1}^{\infty} \frac{{(-1)}^n \tan{(n)}}{n^2}$

Question

Convergence/divergence of $$\sum_{n=1}^{\infty} \frac{{(-1)}^n \tan{(n)}}{n^2}$$ I thought diverge because some value $n \approx \frac{\pi}{2}+\pi k \implies \tan{n}=\pm \infty$ but key says that this never holds because $\pi$ is irrational. He said $\tan{n}$ is bounded and clearly series without $\tan{n}$ converges. Someone can explain this further to me please?

Answer 1

Let $\frac{p_n}{q_n}$ be a convergent of the continued fraction of $\pi$ such that $q_n$ is odd and $p_n$ is even. We have $$ \left|\frac{\pi}{2} q_n - \frac{p_n}{2} \right| \leq \frac{1}{2q_n} $$ and since both $\sin$ and $\cos$ are Lipschitz-continuous we have $$\left|\sin\left(\frac{p_n}{2}\right)\right|\geq 1-\frac{1}{2q_n}\approx 1-\frac{\pi}{4(p_n/2)}, $$ $$\left|\cos\left(\frac{p_n}{2}\right)\right|\leq \frac{1}{2q_n}\approx \frac{\pi}{4(p_n/2)} $$ so there is a sequence of natural numbers $n$ such that $|\tan(n)|$ is at least as large as $\left(\frac{2}{\pi}-\varepsilon\right)n$.
If we assume that the irrationality measure of $\pi$ is $>3$ we have that $\frac{\tan n}{n^2}$ is not even bounded.
On the other hand the irrationality measure of $\pi$ is still unknown (it is conjectured to be $2$, but nowadays we only know that it is $\leq 7.11$), so to discuss the convergence of such series, like the Flint Hills series, is pretty pointless.
Your series is probably convergent since $(-1)^n$ has bounded partial sums and $\frac{\tan(n)}{n^2}$ is probably convergent to zero without wild oscillations, but we currently lack the technology for proving such a claim.