Sum $\sum_{n=1}^{\infty}\frac{1}{n^3}$

Question

Probably you would have known the answer of $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$ I then became curious about the reciprocal of cubes, not squares, like $$\sum_{n=1}^{\infty}\frac{1}{n^3}$$Can you find any answer to this? I have no clue of even where to start.

It doesn't have any closed form and it is known as apery's constant — Aditya Dwivedi, Aug 25 '20 at 09:20
I think that would evaluate to 1.(something). - tried to do some guesswork - — Spectre, Aug 25 '20 at 09:23
In case you're curious, closed forms are obtainable for $\zeta(n)$ when $n>1$ is even, but not when it's odd. This is related to the identity $\sin\pi z=\pi z\prod_{k\ge1}(1-z^2/k^2)$. — J.G., Aug 25 '20 at 09:37
@ArticChar I don't agree with the current duplicate. Maybe this one is more appropriate. — user, Aug 25 '20 at 10:10
@user I agree that the question you linked is the better target and changed the duplicate target accordingly. In similar cases, please flag for moderator attention so that the target change/addition can be done in a timely fashion. — Daniel Fischer, Aug 25 '20 at 19:26
@DanielFischer Thanks for your kind attention and intervention. Regards — user, Aug 25 '20 at 20:05

user · Accepted Answer · 2020-08-25T09:39:30.323

As noticed in the comment its value is known as Apéry's constant which is related to Riemann zeta function, we can estimate its value by a simple integral evaluation

$$\sum_{n=1}^{\infty}\frac{1}{n^3}\approx 1+\frac18+\frac1{27}+\frac1{64}+\frac1{125}+\int_5^\infty \frac{1}{x^3}dx=\\=1+\frac18+\frac1{27}+\frac1{64}+\frac1{125}+\frac1{2\cdot 5^2}\approx 1.20566$$

which can be improved adding more terms

$$\sum_{n=1}^{\infty}\frac{1}{n^3}\approx 1+\frac18+\frac1{27}+\frac1{64}+\frac1{125}+\ldots+\frac1{k^3}+\frac1{2k^2}$$

Sum $\sum_{n=1}^{\infty}\frac{1}{n^3}$

1 Answers1