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I'm curious of what does this sum:

$1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\frac{1}{125}+\frac{1}{216}+...+(\frac{1}{n})^3$

or the Riemann zeta function: $\zeta({3})$ approach. I watched a few 3Blue1Brown videos on YouTube, but it doesn't have any videos about what does that sum above approach. I don't know how to prove it, and I tried, but suddenly all of my tries lead to no results. Is this problem unsolved? If not, how would you prove it?

I-85a
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  • It approaches, by definition, $\zeta(3)\ldots$ There's nothing to prove, except perhaps that the sum converges (obvious by integral test) – Brevan Ellefsen Nov 04 '18 at 22:30

2 Answers2

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$\zeta(3)$ is a constant known as Apery's constant. Its value is approximately $1.2021\ldots$ but as far as I know an exact value isn't known.

Some further reading:

https://en.wikipedia.org/wiki/Ap%C3%A9ry%27s_constant http://mathworld.wolfram.com/AperysConstant.html

PrincessEev
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  • I don't know what it is you think "isn't quite known". The "exact value" is $\zeta(3)$. Apéry proved it is irrational. – Robert Israel Nov 04 '18 at 22:49
  • "The exact value of $\zeta(3)$ is $\zeta(3)$" isn't exactly an enlightening answer. I mean something in terms of - at least more known - constants, i.e. how we know $\zeta(2) = \pi^2/6$. A "closed form" value, perhaps? Whatever the term is. – PrincessEev Nov 04 '18 at 22:51
  • It is not likely that $\zeta(3)$ has a closed form in terms of "better-known" constants, though of course this has not been proven. We don't even have a proof that it is transcendental. – Robert Israel Nov 04 '18 at 23:00
  • Well, there is $\zeta(3) = -\Psi''(1)/2$, but I don't know if you'd count that as "better-known". – Robert Israel Nov 05 '18 at 02:26
  • "The exact value of ζ(3) is ζ(3)" isn't exactly an enlightening answer." Why not? Why would not you say "the exact value of $\pi$ isn't quite known"? – user Nov 05 '18 at 22:07
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Well, $\sum_{n\geq 1}\frac{1}{n^3}$ is a number, precisely the value of the Riemann $\zeta$ function at $s=3$. Since $x\leq \text{arctanh}(x)$ is a pretty tight approximation for $x\to 0^+$,

$$\zeta(3) = 1+\sum_{n\geq 2}\frac{1}{n^3} \leq 1+\frac{1}{2}\sum_{n\geq 2}\log\left(\frac{n^3+1}{n^3-1}\right)$$ but $\prod_{n\geq 2}\frac{n^3+1}{n^3-1}=\prod_{n\geq 2}\frac{n+1}{n-1}\cdot\frac{n^2-n+1}{n^2+n+1}$ is a telescopic product convergent to $\frac{3}{2}$, hence $$ \zeta(3) \approx 1+\frac{\log 3-\log 2}{2}.$$ By creative telescoping we also have the acceleration formula $$ \zeta(3)=\sum_{n\geq 1}\frac{1}{n^3}=\frac{5}{2}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}$$ (see my notes for a proof) which allowed Apery to prove that $\zeta(3)\not\in\mathbb{Q}$.
The irrationality of $\zeta(5),\zeta(7),\zeta(9),\ldots$ still is an open problem, like the conjecture $\zeta(3)\in\pi^3\mathbb{Q}$, which looks numerically very unlikely. On the other hand $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^3}\in\pi^3\mathbb{Q}$, as shown here.

Jack D'Aurizio
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