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What is the sum of $$\sum_{n=3}^\infty \frac{\tan\left(\frac{\pi}{n}\right)}{n}\quad ?$$

I was playing a little bit with geometry things, and I got this, this would be the sum of the inverse areas of all regular polygons of length 1, it's a silly thing, but I'm interested in if there's a way to calculate the exact value of this sum. (Actually the sum have a 4 multiplying, but I guess it is not relevant for the question).

Night
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1 Answers1

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A closed form does not seem to exist but a very close numerical approximation could be obtained $$\sum_{n=3}^\infty \frac{1}{n}\tan\left(\frac{\pi}{n}\right)=\frac{1}{4}+\frac{7}{6 \sqrt{3}}+\frac{\sqrt{5-2 \sqrt{5}}}{5} +\sum_{n=7}^\infty \frac{1}{n}\tan\left(\frac{\pi}{n}\right)$$ Now, we can use the Taylor expansion of the summand to $O\left(\frac{1}{n^{2p+2}}\right)$ to get things like $$\sum_{n=7}^\infty \frac{1}{n}\tan\left(\frac{\pi}{n}\right)=\sum_{n=7}^\infty \left(\frac{\pi }{n^2}+\frac{\pi ^3}{3 n^4}+\frac{2 \pi ^5}{15 n^6}+\frac{17 \pi ^7}{315 n^8}+\frac{62 \pi ^9}{2835 n^{10}}+O\left(\frac{1}{n^{12}}\right)\right) $$ and use the fact that $$\sum_{n=7}^\infty \frac{1}{n^{2k}}=\zeta (2 k,7)$$ where appears the Hurwitz zeta function.

The summation converges quite quickly $$\left( \begin{array}{cc} 10 &\color{red}{ 1.56439140145}772435482499354272938186565292861596865524537235 \\ 20 &\color{red}{ 1.5643914014585097722127}2086003618544015495949712484440355570 \\ 30 &\color{red}{ 1.564391401458509772212730667895116}42880289689323917105514653 \\ 40 &\color{red}{ 1.56439140145850977221273066789511655725847686}518077062210715 \\ 50 &\color{red}{ 1.564391401458509772212730667895116557258476866868456148}18312 \\ 60 &\color{red}{ 1.56439140145850977221273066789511655725847686686845614820529} \end{array} \right)$$ which is not recognized by inverse symbolic calculators.

Edit

If instead of using Taylor series, we use the $[6,2]$ Padé approximant $$\frac{1}{n}\tan\left(\frac{\pi}{n}\right)\sim\frac {\frac{\pi }{n^2}-\frac{\pi ^3}{14 n^4}-\frac{\pi ^5}{630 n^6} } {1-\frac{17 \pi ^2}{42 n^2} }=\frac{\pi ^3}{255 n^4}+\frac{269 \pi }{1445 n^2}+\frac{49392 \pi }{1445 \left(42 n^2-17 \pi ^2\right)}$$ we have $$\sum_{n=7}^\infty \frac{1}{n^{4}}=\frac{\pi ^4}{90}-\frac{14011361}{12960000}$$ $$\sum_{n=7}^\infty \frac{1}{n^{2}}=\frac{\pi ^2}{6}-\frac{5369}{3600}$$ $$\sum_{n=7}^\infty \frac{1}{42 n^2-17 \pi ^2}=\frac{1}{34 \pi ^2}+\frac{1}{17 \pi ^2-1512}+\frac{1}{17 \pi ^2-1050}+\frac{1}{17 \pi ^2-672}+$$ $$\frac{1}{17 \pi ^2-378}+\frac{1}{17 \pi ^2-168}+\frac{1}{17 \pi ^2-42}-\frac{1}{2 \sqrt{714}}\cot \left(\sqrt{\frac{17}{42}} \pi ^2\right)$$ and all of that gives, as an approximation, a closed form which numerically is $1.564391397$

Doing the same for the $[2p,2]$ Padé approximant, we always get closed forms and the numbers are $$\left( \begin{array}{cc} p & \text{result} \\ 1 & 1.5642618568560180372317 \\ 2 & 1.5643907830400816846897 \\ 3 & 1.5643913971007686823770 \\ 4 & 1.5643914014236644621416 \\ 5 & 1.5643914014582161897902 \\ 6 & 1.5643914014585072339117 \\ 7 & 1.5643914014585097499406 \\ 8 & 1.5643914014585097720155 \\ 9 & 1.5643914014585097722109 \\ 10 & 1.5643914014585097722127 \end{array} \right)$$

In all formulae, for sure, appear a term $\cot(\sqrt{a_p} \pi^2)$ where the $a_p$'s make the sequence $$\left\{\frac{1}{3},\frac{2}{5},\frac{17}{42},\frac{62}{153},\frac{691}{1705},\frac{ 10922}{26949},\frac{929569}{2293620},\frac{6404582}{15802673},\frac{221930581}{5 47591761},\frac{9444233042}{23302711005}\right\}$$

Claude Leibovici
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  • Another approach. Cheers! :D – PinkyWay Aug 07 '20 at 10:58
  • Because I am silly, I wrote a test program using quadmath (IEEE Binary128 floating point, with 112 bits of mantissa (33 decimal digits of precision), to brute-force the series (using Kahan summation). After 10,000,000,000 iterations, it had reached 1.564391401 (it does about a million per second on my particular machine), in full agreement with your first results. The approximation is too small. With Binary128, if we assume 1 ULP error per operation, those iterations can incur less than $2^{35}$ ULPs of error, and since the mantissa has 112 bits, there is at least 77 bits of [...] – None Aug 07 '20 at 12:41
  • [...] meaninful information left – over 20 decimal digits. (In practice, more, because some of the rounding loss cancels out; that is just the worst case estimate of error.) This is why I used such an "unreasonable" type and Kahan summation for this brute force test, instead of just standard double: with iteration counts in the billions, the accumulated quantization error (rounding error inherent in the limited-precision floating-point arithmetic used, for every operation from $\tan$ to division and addition) eats away the numerical precision (information content) in the sum mantissa. – None Aug 07 '20 at 12:42
  • @None. Thanks for the information. If you have time to waste, have a look at my most recent edit. Cheers :-) – Claude Leibovici Aug 07 '20 at 13:16
  • @ClaudeLeibovici: Thanks for the edit; I was unfamiliar with Padé approximants (or rather, that approximation with rational functions is called that), and it is only the $[6,2]$ closed form that is too small; both the Taylor series and the $[2p, 2]$ Padé approximations agree with the brute-force numerical approximation – not just numerically, but also in how they converge; obviously just much, much faster than the original series. This is useful to know to me, thanks for your answer! – None Aug 07 '20 at 20:48