What does $1+\frac{1}{8}+\frac{1}{27}+\frac{1}{64}+\frac{1}{125}+(\frac{1}{n})^3$ equal to?

Question

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?

Answer 1

$\zeta(3)$ 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

Answer 2

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.