<s>[INST] zeta of three, question about closed form If $\sum\limits_{n=2}^\infty \frac1{(n^2-n)^3}=10-\pi^2$, then what is the limit in closed form of $\sum\limits_{n=1}^\infty \frac1{n^3}$?
 [/INST] It's called Apéry's constant, $\zeta(3)$, and it doesn't have any known closed form.

Allow me to demonstrate how the first sum doesn't help. Use $(a-b)^3=a^3-3ab(a-b)-b^3$ below:
$$\sum_{n=2}^\infty \frac{1}{(n^2-n)^3}=\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^3=\sum_{n=1}^\infty \color{DarkOrange}{\frac{1}{n^3}}-\frac{3}{\color{Purple}{n(n+1)}}\color{Purple}{\left(\frac{1}{n}-\frac{1}{n+1}\right)}-\color{DarkOrange}{\frac{1}{(n+1)^3}}.$$
The left and rightmost parts of the summand form a $\rm\color{DarkOrange}{telescoping}$ series, so we evaluate this to $1$ and continue on with our derivation with further partial fraction decomposition:
$$=\color{DarkOrange}{1}-3\sum_{n=1}^\infty \color{Purple}{\left(\frac{1}{n}-\frac{1}{n+1}\right)^2}=
1-3\sum_{n=1}^\infty\left(\color{Red}{\frac{1}{n^2}}-\color{Green}{\frac{2}{n(n+1)}}+\color{Blue}{\frac{1}{(n+1)^2}}\right)$$
This time the left and rightmost parts are almost identical copies of $\zeta(2)$ and the middle telescoping:
$$=1-3\big(\color{Red}{\zeta(2)}-\color{Green}2+\color{Blue}{\zeta(2)-1}\big)=10-\pi^2.$$
As you can see, the copies of $\color{DarkOrange}{\zeta(3)}$ cancelled each other out while the $\zeta(2)$s didn't.
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