Consider the infinite series $$ F(s)=\sum_{k=1}^\infty \frac{1}{k^{2s+1} \sin(k \pi \sqrt{2})} $$ Prudnikov Vol.1 , gives a result for this sum in 5.4.16.13 for $s=1.$ $$ F(1)=-\frac{13 \pi^3}{360 \sqrt{2}} $$ but no reference. How do you prove this result?
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3$\begingroup$ The article "ON THE CONVERGENCE OF DIOPHANTINE DIRICHLET SERIES" (Proceedings of the Edinburgh Mathematical Society (2012) 55, 513–541) by T. Rivoal adresses generalizations of this kind of sums. Apparently, these was (first?) considered by Hardy and Littlewood in their article "Some problems of Diophantine approximation: a series of cosecant". $\endgroup$– efsCommented Nov 17, 2021 at 4:24
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3$\begingroup$ Rivoal's article is here $\endgroup$– David Roberts ♦Commented Nov 17, 2021 at 6:20
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1$\begingroup$ The series definitely converges, since the distance from $k\sqrt{2}$ to the closest integer is not less than $c/k$ $\endgroup$– Fedor PetrovCommented Nov 17, 2021 at 6:52
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5$\begingroup$ this is answered on math.stackexchange.com/a/1658310/87355 ; it follows upon evaluation of the contour integral $$\oint_{C}\frac{dz}{z^3[\sin(\pi z)\sin(\sqrt{2}-1)\pi z]},\;\; C=\text{square with vertices at}\;\pm (N-1/2) (1 \pm i),$$ which vanishes in the limit $N\rightarrow\infty$. The residues contain the desired sum over hyperbolic functions. $\endgroup$– Carlo BeenakkerCommented Nov 17, 2021 at 7:16
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6$\begingroup$ "inverse trig function" means $\arcsin x$, not $1/\sin x$. $\endgroup$– Gerry MyersonCommented Nov 17, 2021 at 8:57
1 Answer
Thanks Carlo for pointing out the solution @ math.stackexchange from Ron Gordon. Yes, this works. Indeed, one can use the same technique to solve my series for arbitrary positive integer $s$.
For this, start with the the contour pointed out by Carlo above, which vanishes: $$ \oint_{C} dz \frac{1}{z^{2s+1} \sin(\pi z)\sin(\pi z (\sqrt{2}-1))}=0\,. $$ The contour has simple poles for all integers except zero: $z=n$, $n=\pm 1,\pm 2,\pm 3,\ldots$; simple pose for all integers times $\sqrt{2}+1$: $z=(\sqrt{2}+1)n$ with the same n's and finally a pole of order $2s+3$ at $z=0$.
The residues of the poles near $z=n$ are found to be given by $$ Res_{z=n}=\frac{(\sqrt{2}-1)^{2s}}{n^{2s+1}\pi \sin{\pi n \sqrt{2}}}. $$ The residues of the poles near $z=n (\sqrt{2}+1)$ are given by $$ Res_{z=n(\sqrt{2}+1)}=\frac{(\sqrt{2}-1)^{2s}}{n^{2s+1}\pi \sin(\pi n \sqrt{2})}. $$
To find the residue of the pole at $z=0$, one can use the expansion of the cosecans function in terms of Bernoulli numbers: $$ \csc(\pi x)=\sum_{n=0}^\infty \frac{B_{2n} (2^{2n}-2) (-1)^{n+1}}{(2n)!} (\pi x)^{2n-1}, $$ for both cosecans functions in the denominator. Here $B_{2n}$ are the Bernoulli numbers. The residue is the coefficient of the $1/z$ term in the resulting double sum, which is then given by $$ Res_{z=0}=\sum_{n=0}^{s+1} \frac{(-1)^{s+1} (2^{2n}-2)(2^{2n-2s+2}-2) B_{2n} B_{2s-2n+2} \pi^{2s} (\sqrt{2}-1)^{2n-1}}{(2n)!(2s-2n+2)!}. $$ Since the contour integral vanishes, the sums of the residue must also vanish, and one finds $$ \sum_{n=1}^\infty \frac{1}{n^{2s+1}\sin (n \pi \sqrt{2})}=\frac{\pi^{2s+1} (-1)^{s}}{2 (1+(\sqrt{2}-1)^{2s})} \sum_{n=0}^{s+1} \frac{(2^{2n}-2)(2^{2s-2n+2}-2) B_{2n}B_{2s-2n+2} (\sqrt{2}-1)^{2n-1}}{(2n)!(2s-2n+2)!}, $$ proving Prudnikov's equation 5.4.16.12. Using the relation of the Bernoulli numbers and the Riemann zeta function, one can rewrite this as $$ \sum_{n=1}^\infty \frac{1}{n^{2s+1}\sin (n \pi \sqrt{2})}=-\frac{2^{-2s-1}}{\pi (1+(\sqrt{2}-1)^{2s})} \sum_{n=0}^\infty \zeta(2n)\zeta(2s-2n+2) (2^{2n}-2)(2^{2s-2n+2}-2) (\sqrt{2}-1)^{2n-1}. $$ I wonder if this could be used to define the sum away from positive integer $s$.