Classical

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]

Fast convergence

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]

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

Digit by digit

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]

Thue-Morse sequence

The following representation was found by Tóth in 2022:^[16]

${\displaystyle {\begin{aligned}\sum _{n\geq 1}{\frac {9t_{n-1}+7t_{n}}{n^{3}}}&=8\zeta (3),\end{aligned}}}$

where ${\displaystyle (t_{n})_{n\geq 0}}$ is the ${\displaystyle n^{\rm {th}}}$ term of the Thue-Morse sequence. In fact, this is a special case of the following formula (valid for all ${\displaystyle s}$ with real part greater than ${\displaystyle 1}$):

${\displaystyle (2^{s}+1)\sum _{n\geq 1}{\frac {t_{n-1}}{n^{s}}}+(2^{s}-1)\sum _{n\geq 1}{\frac {t_{n}}{n^{s}}}=2^{s}\zeta (s).}$

Others

The following series representation was found by Ramanujan:^[17]

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

The following series representation was found by Simon Plouffe in 1998:^[18]

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

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

Simple formulas

The following formula follows directly from the integral definition of the zeta function:

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

More complicated formulas

Other formulas include^[19]

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

and^[20]

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

Also,^[21]

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

A connection to the derivatives of the gamma function^[22]

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

is also very useful for the derivation of various integral representations via the known integral formulas for the gamma and polygamma functions.^[23]