TITLE: Does the sum $\sum_{n=1}^{\infty}\frac{\tan n}{n^2}$ converge?
QUESTION [39 upvotes]: Does the sum $$\sum_{n=1}^{\infty}\frac{\tan n}{n^2}$$
converge?

REPLY [2 votes]: note
$$
\cos(x) = \prod_{n=1}^{\infty}\left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right)
$$
then
$$
\log(\cos(x)) = \sum_{n=1}^{\infty}\log\left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right)
$$
which gives
$$
=-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{4^k x^{2k}}{\pi^{2k}{k} \ {(2n-1)}^{2k}}
$$
summing over n yields
$$
=-\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^k x^{2k}}{\pi^{2k}{k}}
$$
where $ \xi(s) = \zeta(s)\left(1 - \frac{1}{2^s}\right)$ and $\zeta(s)$ is the Riemann Zeta Function.  Derivation on both sides with respect to $x$ yields
$$
-\frac{\sin(x)}{\cos(x)} = -2\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^kx^{2k-1}}{\pi^{2k}}
$$
Cancelling the negative, dividing by $x^2$, and summing over $x$ gives
$$
\sum_{x=1}^{\infty}\frac{\tan(x)}{x^2} = 2\sum_{x=1}^{\infty}\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^k x^{2k-3}}{\pi^{2k}}=2\sum_{k=1}^{\infty}\frac{{\xi(2k)} \zeta(3-2k)\ 4^{k}}{\pi^{2k}}
$$ 
At $k = 1$ we have $\zeta(3 - 2(1)) = \zeta(1) = \infty \therefore$ the R.H.S must diverge.