Apéry's constant

In addition to the fundamental series:

${\displaystyle \zeta (3)=\sum _{k=1}^{\infty }{\frac {1}{k^{3}}},}$

Leonhard Euler gave the series representation:[6]

${\displaystyle \zeta (3)={\frac {\pi ^{2}}{7}}\left(1-4\sum _{k=1}^{\infty }{\frac {\zeta (2k)}{2^{2k}(2k+1)(2k+2)}}\right)}$

in 1772, which was subsequently rediscovered several times.[7]

Other classical series representations include:

${\displaystyle {\begin{aligned}\zeta (3)&={\frac {8}{7}}\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{3}}}\\\zeta (3)&={\frac {4}{3}}\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(k+1)^{3}}}\end{aligned}}}$

Since the 19th century, a number of mathematicians have found convergence acceleration series for calculating decimal places of ζ(3). Since the 1990s, this search has focused on computationally efficient series with fast convergence rates (see section "Known digits").

The following series representation was found by A. A. Markov in 1890,[8] rediscovered by Hjortnaes in 1953,[9] and rediscovered once more and widely advertised by Apéry in 1979:[3]

${\displaystyle \zeta (3)={\frac {5}{2}}\sum _{k=1}^{\infty }(-1)^{k-1}{\frac {k!^{2}}{(2k)!k^{3}}}}$

The following series representation gives (asymptotically) 1.43 new correct decimal places per term:[10]

${\displaystyle \zeta (3)={\frac {1}{4}}\sum _{k=1}^{\infty }(-1)^{k-1}{\frac {(k-1)!^{3}(56k^{2}-32k+5)}{(2k-1)^{2}(3k)!}}}$

The following series representation gives (asymptotically) 3.01 new correct decimal places per term:[11]

${\displaystyle \zeta (3)={\frac {1}{64}}\sum _{k=0}^{\infty }(-1)^{k}{\frac {k!^{10}(205k^{2}+250k+77)}{(2k+1)!^{5}}}}$

The following series representation gives (asymptotically) 5.04 new correct decimal places per term:[12]

${\displaystyle \zeta (3)={\frac {1}{24}}\sum _{k=0}^{\infty }(-1)^{k}{\frac {(2k+1)!^{3}(2k)!^{3}k!^{3}(126392k^{5}+412708k^{4}+531578k^{3}+336367k^{2}+104000k+12463)}{(3k+2)!(4k+3)!^{3}}}}$

It has been used to calculate Apéry's constant with several million correct decimal places.[13]

The following series representation gives (asymptotically) 3.92 new correct decimal places per term:[14]

${\displaystyle \zeta (3)={\frac {1}{2}}\,\sum _{k=0}^{\infty }{\frac {(-1)^{k}(2k)!^{3}(k+1)!^{6}(40885k^{5}+124346k^{4}+150160k^{3}+89888k^{2}+26629k+3116)}{(k+1)^{2}(3k+3)!^{4}}}}$

In 1998, Broadhurst gave a series representation that allows arbitrary binary digits to be computed, and thus, for the constant to be obtained in nearly linear time, and logarithmic space.[15]

The following series representation was found by Ramanujan:[16]

${\displaystyle \zeta (3)={\frac {7}{180}}\pi ^{3}-2\sum _{k=1}^{\infty }{\frac {1}{k^{3}(e^{2\pi k}-1)}}}$

The following series representation was found by Simon Plouffe in 1998:[17]

${\displaystyle \zeta (3)=14\sum _{k=1}^{\infty }{\frac {1}{k^{3}\sinh(\pi k)}}-{\frac {11}{2}}\sum _{k=1}^{\infty }{\frac {1}{k^{3}(e^{2\pi k}-1)}}-{\frac {7}{2}}\sum _{k=1}^{\infty }{\frac {1}{k^{3}(e^{2\pi k}+1)}}.}$

Srivastava (2000) collected many series that converge to Apéry's constant.

For example, this one follows from the summation representation for Apéry's constant:

${\displaystyle \zeta (3)=\int _{0}^{1}\!\!\int _{0}^{1}\!\!\int _{0}^{1}{\frac {1}{1-xyz}}\,dx\,dy\,dz}$.

The next two follow directly from the well-known integral formulas for the Riemann zeta function:

${\displaystyle \zeta (3)={\frac {1}{2}}\int _{0}^{\infty }{\frac {x^{2}}{e^{x}-1}}\,dx}$

and

${\displaystyle \zeta (3)={\frac {2}{3}}\int _{0}^{\infty }{\frac {x^{2}}{e^{x}+1}}\,dx}$.

This one follows from a Taylor expansion of χ₃(e^ix) about x = ±π/2, where χ_ν(z) is the Legendre chi function:

${\displaystyle \zeta (3)={\frac {4}{7}}\int _{0}^{\frac {\pi }{2}}x\log {(\sec {x}+\tan {x})}\,dx}$

Note the similarity to

${\displaystyle G={\frac {1}{2}}\int _{0}^{\frac {\pi }{2}}\log {(\sec {x}+\tan {x})}\,dx}$

where G is Catalan's constant.

Other formulas include:[18]

${\displaystyle \zeta (3)=\pi \!\!\int _{0}^{\infty }\!{\frac {\cos(2\arctan {x})}{\left(x^{2}+1\right)\left(\cosh {\frac {1}{2}}\pi x\right)^{2}}}\,dx}$,

and,[19]

${\displaystyle \zeta (3)=-{\frac {1}{2}}\int _{0}^{1}\!\!\int _{0}^{1}{\frac {\log(xy)}{\,1-xy\,}}\,dx\,dy=-\int _{0}^{1}\!\!\int _{0}^{1}{\frac {\log(1-xy)}{\,xy\,}}\,dx\,dy}$,

Mixing these two formulas, one can obtain :

${\displaystyle \zeta (3)=\int _{0}^{1}\!\!{\frac {\log(x)\log(1-x)}{\,x\,}}\,dx}$,

By symmetry,

${\displaystyle \zeta (3)=\int _{0}^{1}\!\!{\frac {\log(x)\log(1-x)}{\,1-x\,}}\,dx}$,

Summing both, ${\displaystyle \zeta (3)={\frac {1}{2}}\int _{0}^{1}\!\!{\frac {\log(x)\log(1-x)}{\,x(1-x)\,}}\,dx}$.

Also,[20]

${\displaystyle {\begin{aligned}\zeta (3)&={\frac {8\pi ^{2}}{7}}\!\!\int _{0}^{1}\!{\frac {x\left(x^{4}-4x^{2}+1\right)\log \log {\frac {1}{x}}}{\,(1+x^{2})^{4}\,}}\,dx\\&={\frac {8\pi ^{2}}{7}}\!\!\int _{1}^{\infty }\!{\frac {x\left(x^{4}-4x^{2}+1\right)\log \log {x}}{\,(1+x^{2})^{4}\,}}\,dx\end{aligned}}}$.

A connection to the derivatives of the gamma function

${\displaystyle \zeta (3)=-{\tfrac {1}{2}}\Gamma '''(1)+{\tfrac {3}{2}}\Gamma '(1)\Gamma ''(1)-{\big (}\Gamma '(1){\big )}^{3}=-{\tfrac {1}{2}}\,\psi ^{(2)}(1)}$

is also very useful for the derivation of various integral representations via the known integral formulas for the gamma and polygamma-functions.[21]