This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

affil0affil0affiliationtext: Hürth, Germany

Recurrence Relations for Values of the Riemann Zeta Function in Odd Integers

Tobias Kyrion
Abstract

It is commonly known that ζ(2k)=qkζ(2k+2)π2\zeta(2k)=q_{k}\frac{\zeta(2k+2)}{\pi^{2}} with known rational numbers qkq_{k}. In this work we construct recurrence relations of the form k=1rkζ(2k+1)π2k=0\sum_{k=1}^{\infty}r_{k}\frac{\zeta(2k+1)}{\pi^{2k}}=0 and show that series representations for the coefficients rkr_{k}\in\mathbb{R} can be computed explicitly.

1 Summary

In the first section we show that cosh(x)sinh(x)2N+1\frac{\cosh(x)}{\sinh(x)^{2N+1}} can be expressed as linear combination of sinh(x)cosh(x)2N+1\frac{\sinh(x)}{\cosh(x)^{2N+1}} and cosh(2x)sinh(2x)2k+1\frac{\cosh(2x)}{\sinh(2x)^{2k+1}} for some kNk\leq N. We achieve this by proving four identities between certain rational functions. Then we show that the 2n2n-th derivative of coth(x)\coth(x) can be expressed as linear combination of cosh(x)sinh(x)2k+1\frac{\cosh(x)}{\sinh(x)^{2k+1}} with kk ranging from 11 to nn. We prove some useful recurrence relations between the coefficients of the cosh(x)sinh(x)2k+1\frac{\cosh(x)}{\sinh(x)^{2k+1}}’s and compute explicitly the inverse of the matrix formed by these coefficients. We derive our main result Theorem 3.3 - the limit identity for limα0+n=11nsinh(αn)cosh(αn)2N+1\lim_{\alpha\to 0_{+}}\sum_{n=1}^{\infty}\frac{1}{n}\frac{\sinh(\alpha n)}{\cosh(\alpha n)^{2N+1}} for a fixed NN\in\mathbb{N} - by applying our previous findings on Ramanujan’s famous identity for the Riemann zeta function values in odd integers. As an application we finally determine recurrence relations of the form k=1rkζ(2k+1)π2k=0\sum_{k=1}^{\infty}r_{k}\frac{\zeta(2k+1)}{\pi^{2k}}=0.

2 Preliminaries

Remark 2.1.

Throughout this work we set (nk)=0\binom{n}{k}=0 for k<0k<0 and for n<kn<k.

2.1 Identities for cosh(x)sinh(x)2N+1\frac{\cosh(x)}{\sinh(x)^{2N+1}}

Proposition 2.2.

Let MM\in\mathbb{N} and zz\in\mathbb{C}. Then we have the following four relations between rational functions:

12(z+1z)4M+212(z1z)4M+2\displaystyle\frac{1}{2}\left(z+\frac{1}{z}\right)^{4M+2}-\frac{1}{2}\left(z-\frac{1}{z}\right)^{4M+2} =k=0M24k4M+22k+1(M+k2k)(z21z2)2(Mk)\displaystyle=\;\sum_{k=0}^{M}2^{4k}\frac{4M+2}{2k+1}\binom{M+k}{2k}\left(z^{2}-\frac{1}{z^{2}}\right)^{2(M-k)}
12(z+1z)4M12(z1z)4M\displaystyle\frac{1}{2}\left(z+\frac{1}{z}\right)^{4M}-\frac{1}{2}\left(z-\frac{1}{z}\right)^{4M} = 4(z2+1z2)k=0M124kMk2k+1(M+k2k)(z21z2)2(M1k)\displaystyle=\;4\left(z^{2}+\frac{1}{z^{2}}\right)\sum_{k=0}^{M-1}2^{4k}\frac{M-k}{2k+1}\binom{M+k}{2k}\left(z^{2}-\frac{1}{z^{2}}\right)^{2(M-1-k)}
12(z+1z)4M+12(z1z)4M\displaystyle\frac{1}{2}\left(z+\frac{1}{z}\right)^{4M}+\frac{1}{2}\left(z-\frac{1}{z}\right)^{4M} =k=0M24kMM+k(M+k2k)(z21z2)2(Mk)\displaystyle=\;\sum_{k=0}^{M}2^{4k}\frac{M}{M+k}\binom{M+k}{2k}\left(z^{2}-\frac{1}{z^{2}}\right)^{2(M-k)}
12(z+1z)4M+2+12(z1z)4M+2\displaystyle\frac{1}{2}\left(z+\frac{1}{z}\right)^{4M+2}+\frac{1}{2}\left(z-\frac{1}{z}\right)^{4M+2} =(z2+1z2)k=0M24k(M+k2k)(z21z2)2(Mk).\displaystyle=\;\left(z^{2}+\frac{1}{z^{2}}\right)\sum_{k=0}^{M}2^{4k}\binom{M+k}{2k}\left(z^{2}-\frac{1}{z^{2}}\right)^{2(M-k)}.
Proof.

After factoring out z4(Mj)+z4(jM)z^{4(M-j)}+z^{4(j-M)} in both sides of the first relation and z4(Mj)2+z4(jM)+2z^{4(M-j)-2}+z^{4(j-M)+2} in the second, z4(Mj)+z4(jM)z^{4(M-j)}+z^{4(j-M)} in the third and z4(Mj)+2+z4(jM)2z^{4(M-j)+2}+z^{4(j-M)-2} in the fourth we are by comparing coefficients left to show the following relations:

k=0j(1)jk24k((2(Mk)jk)(2(Mk)j1k))(M+k2k)\displaystyle\sum_{k=0}^{j}(-1)^{j-k}2^{4k}\left(\binom{2(M-k)}{j-k}-\binom{2(M-k)}{j-1-k}\right)\binom{M+k}{2k} =(4M+22j),\displaystyle=\binom{4M+2}{2j}, j\displaystyle j =0,,M\displaystyle=0,...,M
k=0j(1)jk24kMM+k(2(Mk)jk)(M+k2k)\displaystyle\sum_{k=0}^{j}(-1)^{j-k}2^{4k}\frac{M}{M+k}\binom{2(M-k)}{j-k}\binom{M+k}{2k} =(4M2j),\displaystyle=\binom{4M}{2j}, j\displaystyle j =0,,M\displaystyle=0,...,M
k=0j(1)jk24k4M+22k+1(2(Mk)jk)(M+k2k)\displaystyle\sum_{k=0}^{j}(-1)^{j-k}2^{4k}\frac{4M+2}{2k+1}\binom{2(M-k)}{j-k}\binom{M+k}{2k} =(4M+22j+1),\displaystyle=\binom{4M+2}{2j+1}, j\displaystyle j =0,,M\displaystyle=0,...,M
k=0j(1)jk24k+2Mk2k+1((2(M1k)jk)(2(M1k)j1k))(M+k2k)\displaystyle\sum_{k=0}^{j}(-1)^{j-k}2^{4k+2}\frac{M-k}{2k+1}\left(\binom{2(M-1-k)}{j-k}-\binom{2(M-1-k)}{j-1-k}\right)\binom{M+k}{2k} =(4M2j+1),\displaystyle=\binom{4M}{2j+1}, j\displaystyle j =0,,M1.\displaystyle=0,...,M-1.

We denote with S2j4M+2S_{2j}^{4M+2}, S2j4MS_{2j}^{4M}, S2j+14M+2S_{2j+1}^{4M+2} and S2j+14MS_{2j+1}^{4M} the left hand sides of above relations. With some algebra we can establish

(2M2j+3)(4M+2)jS2j4M+2+4(2M2j+2)\displaystyle(2M-2j+3)(4M+2)jS_{2j}^{4M+2}+4(2M-2j+2) (2M2j+1)(3Mj+2)S2(j1)+14M+2\displaystyle(2M-2j+1)(3M-j+2)S_{2(j-1)+1}^{4M+2}
  2(2M2j+2)(4M+1)(4M+2)S2(j1)+14M\displaystyle-\;\;2(2M-2j+2)(4M+1)(4M+2)S_{2(j-1)+1}^{4M} =(2M2j+1)(2Mj+2)(4M+2)S2(j1)4M+2\displaystyle=-(2M-2j+1)(2M-j+2)(4M+2)S_{2(j-1)}^{4M+2}
jS2j4M2MS2(j1)+14M\displaystyle jS_{2j}^{4M}-2MS_{2(j-1)+1}^{4M} =(2Mj+1)S2(j1)4M\displaystyle=-(2M-j+1)S_{2(j-1)}^{4M}
(2j+1)S2j+14M+2(4M+2)S2j4M+2\displaystyle(2j+1)S_{2j+1}^{4M+2}-(4M+2)S_{2j}^{4M+2} =(4M2j+3)S2(j1)+14M+2\displaystyle=-(4M-2j+3)S_{2(j-1)+1}^{4M+2}
(2M+1)S2j+14M(2M2j)S2j+14M+2\displaystyle(2M+1)S_{2j+1}^{4M}-(2M-2j)S_{2j+1}^{4M+2} =(2M+1)S2(j1)+14M.\displaystyle=(2M+1)S_{2(j-1)+1}^{4M}.

A computation shows that these relations are also fulfilled by the corresponding right hand sides of the top most equations. Thus the proof follows by induction over jj.

\square

Setting z=exp(x)z=\exp(x) for xx\in\mathbb{C} and some algebraic manipulations give