Sum of $\sum _{n=1}^{\infty } \frac{3 n^5 \left(6596 n^8+924 n^4-2856\right)}{e^{2 \pi n}-1}$

Question

I see a puzzle in the picture below.

It seems that the sum $$\sum _{n=1}^{\infty } \frac{3 n^5 \left(6596 n^8+924 n^4-2856\right)}{e^{2 \pi n}-1}$$ is an integer and can form a date. However, even Mathematica cannot give a precise result. The partial sum suggests the result might be 818. How to calculate it?

Interestingly, $$\sum_{n = 1}^{\infty} \frac{3n^5 (2856)}{e^{2 \pi n} - 1} \approx 17$$ seems to be an integer as well. — L. F., Aug 19 '22 at 06:05
In fact, it seems that \begin{align} \sum_{n = 1}^{\infty} \frac{3n^5 (6596n^8)}{e^{2 \pi n} - 1} &\approx 824.5, \ \sum_{n = 1}^{\infty} \frac{3n^5 (924n^4)}{e^{2 \pi n} - 1} &\approx 10.5, \ \sum_{n = 1}^{\infty} \frac{3n^5 (2856)}{e^{2 \pi n} - 1} &\approx 17. \end{align} I suspect that there is a way to calculate $$\sum_{n = 1}^{\infty} \frac{n^k}{e^{2 \pi n} - 1},$$ resulting in rational numbers. — L. F., Aug 19 '22 at 06:11
@L.F. I was about typing the same. It seems to be true for 100 decimals — Claude Leibovici, Aug 19 '22 at 06:17

Claude Leibovici · Answer 1 · 2022-08-19T07:53:14.823

If $$a_n=\frac{3 n^5 \left(6596 n^8+924 n^4-2856\right)}{e^{2 \pi n}-1}$$ the ratio test $$\frac{a_{n+1}}{a_n}=\frac{e^{2 \pi n}-1}{e^{2 \pi (n+1)}-1}\Bigg[1+\frac{13}{n}+\frac{78}{n^2}+O\left(\frac{1}{n^3}\right)\Bigg]\sim e^{-2 \pi }$$ shows an extrely fast convergence.

If we consider $$b_n=\frac{3 n^5 \left(6596 n^8+924 n^4-2856\right)}{e^{2 \pi n}} < a_n$$ we can explicitely compute the infinite sum since $$\sum_{n=1}^\infty \frac {n^k}{e^{2 \pi n}}=\text{Li}_{-k}\left(e^{-2 \pi }\right)$$

$$\sum_{n=1}^\infty b_n=\frac{24 x}{(x-1)^{14}} P_{12}(x)$$ where $x=e^{2\pi}$. I shall not type here the polynomial of degree $12$ in $x$ (just because of the size of the coefficients (the smallest is $583$ but the largest is $1875908773452$; their mean is $394936819200$).

Numerically $$\sum_{n=1}^\infty b_n=817.9491271\cdots$$ Now, consider $$\frac{1}{e^{2 \pi n}-1 }=\sum_{q=0}^\infty e^{-2 \pi (q+1) n}$$ and $$\sum_{n=1}^\infty n^k e^{-2 \pi n (q+1)}=\text{Li}_{-k}\left(e^{-2 \pi (q+1)}\right)$$

Now, the computation of the partial sums $$S_r=\sum_{n=1}^r a_n$$ becomes simple $$\left( \begin{array}{cc} r & S_r \\ 5 & 817.98888429499578340114316254727116027126495 \\ 10 & 817.99999999999933816348471123533611149002402 \\ 15 & 817.99999999999999999999999804247971649003791 \\ 20 & 817.99999999999999999999999999999999999847618 \end{array} \right)$$

Computing $$S_{100}=818-5.616\times 10^{-246}$$ $$S_{1000}=818-6.628\times 10^{-2689}$$

$818$ is the asymptotoic limit of the summation (it will never reach it).

Edit

It seems that I have been $\large\color{red}{\text{wrong}}$. Using Mathematica, type

S[p_]:=NSum[(3*n^5*(-2856 + 924*n^4 + 6596*n^8))/(-1 + E^(2*n*Pi)), {n, 1, Infinity}, WorkingPrecision -> p+3,PrecisionGoal -> p+3, AccuracyGoal -> p+3, NSumTerms -> p]

The result is always $818.$ followed by $p$ trailing $0$'s.

Sum of $\sum _{n=1}^{\infty } \frac{3 n^5 \left(6596 n^8+924 n^4-2856\right)}{e^{2 \pi n}-1}$

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