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

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    $\begingroup$ see also math.stackexchange.com/questions/159438/… $\endgroup$ – al-Hwarizmi Aug 18 '13 at 15:35
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    $\begingroup$ @al-Hwarizmi Note that the paper by Coskey citeseerx.ist.psu.edu/viewdoc/… which is linked in the parallel question deals as its final case with the terms in this sequence (having denominator $n^2$). The final comment is "Thus, the best that Rosenholtz can say is that as far as modern hardware is concerned, the tan(n)/n^2 series stays very small for a very long time." Which suggests the problem is hard. With denominator $n^8$ - the other case discussed - the terms do go to zero. $\endgroup$ – Mark Bennet Aug 18 '13 at 21:07
  • $\begingroup$ yes, right, the paper is ultimate reference that this is not a trivial problem. $\endgroup$ – al-Hwarizmi Aug 19 '13 at 6:38
  • $\begingroup$ @al-Hwarizmi Very interesting Note! Thanks for the link. $\endgroup$ – user119228 Jan 11 '14 at 10:25


$$ \cos(x) = \prod_{n=1}^{\infty}\left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right) $$

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.

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    $\begingroup$ Thank you! But I think $$\frac{\sin(x)}{\cos(x)} = 2\sum_{k=1}^{\infty}\frac{{\xi(2k)}\ 4^kx^{2k-1}}{\pi^{2k}}$$ is not true for $x>\pi/2,$ since the R.H.S must diverge. $\endgroup$ – Next Mar 13 '14 at 16:35

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.

  • $\begingroup$ Someone posted this method in a highly similar question in another thread. I think someone linked that thread (which has the paper you reference) in one of the comments attached to the OP. $\endgroup$ – Alfred Yerger Feb 27 '14 at 17:41

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