Does the sum $\sum_{n=1}^{\infty}\frac{\tan n}{n^2}$ converge?

Solution 1:

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.

Solution 2:

It may be useful to compare two other, similar, series. Mathworld gives definitions and analysis of the Cookson Hill and Flint Hill series, defined as the infinite sum of $\frac{\sec^2 n}{n^3}$ and $\frac{\csc^2 n}{n^3}$, respectively. The article for the Flint Hill series links to a paper by M. A. Alekseyev at arxiv.org, who demonstrates the the general convergence of such series can be linked to the irrationality measure of $\pi$. Giving it a quick look, this might be helpful to sketch out a proof- one could bound tan(x) with other trigonometric functions and then determine what implications such a bounding would have on $\mu(\pi)$, and then determine whether or not this satisfies the current best bounding (around 7.5, IIRC), in which case the series converges, or whether it violates it, in which case the problem is undetermined right now.