A closed form for infinite series $ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$

Question

It is well known that $$ \sum _1 ^\infty \frac {1}{n^2} = \frac {\pi ^2}{6}$$ and $$ \sum _1 ^\infty \frac {1}{n^4} = \frac {\pi ^4}{90}$$

We also know that $$ \sum _1 ^\infty \frac {1}{n^3} $$

$$=1.202056903159594285399738161511449990764986292340498881.....$$

My question is:

Do we have a closed form for this series besides $$ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$$

https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant gives several fast converging infinite sums — Henry, Apr 08 '18 at 10:05
Last time I checked, no, there is no closed form for $\zeta(n)$ where $n \in 2\mathbb{N}+1$. — Rócherz, Apr 08 '18 at 10:18
@MohammadRiazi-Kermani Quoting from Establishing zeta(3) as a definite integral and its computation on MO: "it is conjectured (but not known) that the values $\zeta(2n+1)$ are ... algebraically independent of one another, and of $\pi$.". — dxiv, Apr 09 '18 at 03:33

robjohn · Accepted Answer · 2018-04-09T07:48:50.137

Apéry's Constant

As mentioned in comments $$ \sum_{n=1}^\infty\frac1{n^3}=\zeta(3)\tag1 $$ is also known as Apéry's Constant. There is no closed form in terms of a rational multiple of an integer power of $\pi$.

We also have $$ \sum_{n=1}^\infty\frac1{(2n-1)^3}=\frac78\zeta(3)\tag2 $$

However, as shown in this answer, if we alternate the series in $(2)$, we get $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}\tag3 $$ which is also known as $\beta(3)$, the Dirichlet beta function.

Computation of $\boldsymbol{\zeta(3)}$

The series in $(1)$ converges very slowly. To get the sum to about $10$ places, we would need to sum $100000$ terms.

If we use the Euler-Maclaurin Sum Formula, we get that $$ \sum_{k=1}^n\frac1{k^3}=\small\zeta(3)-\frac1{2n^2}+\frac1{2n^3}-\frac1{4n^4}+\frac1{12n^6}-\frac1{12n^8}+\frac3{20n^{10}}-\frac5{12n^{12}}+O\!\left(\frac1{n^{14}}\right)\tag4 $$ Using $n=10$ in $(4)$, we get $$ \sum_{k=1}^\infty\frac1{k^3}=1.2020569031596\tag5 $$

Thanks for the information. Equation (3) in particular may be of further interest. — Mohammad Riazi-Kermani, Apr 09 '18 at 11:04

score 2 · Answer 2 · answered Dec 13 '18 at 09:33

In addition to robojohn's answer, there are some formulas expressing $\zeta(3)$ (and other odd zeta values) in terms of powers of $\pi$, but these are not clsoed forms. The most well known ones are due to Plouffe and Borwein & Bradley:

$$ \begin{aligned} \zeta(3)&=\frac{7\pi^3}{180}-2\sum_{n=1}^\infty \frac{1}{n^3(e^{2\pi n}-1)},\\ \sum_{n=1}^\infty \frac{1}{n^3\,\binom {2n}n} &= -\frac{4}{3}\,\zeta(3)+\frac{\pi\sqrt{3}}{2\cdot 3^2}\,\left(\zeta(2, \tfrac{1}{3})-\zeta(2,\tfrac{2}{3}) \right). \end{aligned} $$

Moreover, in this Math.SE post we have:

$$ \frac{3}{2}\,\zeta(3) = \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}. $$

You can also check out this paper by Vepstas, which provides a nice generalization to some of these identities.

A closed form for infinite series $ \sum _1 ^\infty \frac {1}{n^3} = \zeta (3) ?$

2 Answers2