$$\sum_{n=1}^\infty \frac{1}{n^3}=?$$
I am a high school student and I want it to answered in a way that I can understand it. I know basic calculus and I will understand it if calculus is used. I also have knowledge of Taylor series and trigonometry and a bit extra of algebra.
Your sum is the value of the Riemann Zeta function at the positive odd integer $3$. If we denote this function by $\zeta(n)$, we have:
$$ \zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}, $$
which converges for all complex numbers $s$ with real part greater than $1$. Many well-established results are known concerning its value at the even positive integers, but little to nothing is known about its value at odd positive integers.
Roger Apéry proved in 1978 that $\zeta(3)=\displaystyle \sum_{n=1}^{\infty} n^{-3}$ is irrational, however no closed form for its value is currently known. He famously proved that
$$ \zeta(3)=\frac52\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^3 {2n \choose n}}, $$
although this formula was already found by Hjortnaes previously, but not widely known at that time.
There are some other formulas that express $\zeta(3)$ (and other odd zeta values) in terms of powers of $\pi$, but these are not closed 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.
