# 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 $\sum\limits_{n=1}^\infty \frac1{n^3}$?

## 2 Answers

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.

• @anon.Since we know the limit of the biggest part of this function the rest it should be some power of pi with a coefficient is this correct? That is the way it works for the second power. – Vassilis Parassidis Feb 9 '12 at 5:10
• @Vassilis: I don't know what you refer to by the "biggest part" versus "the rest" but I assume by "function" you mean "series" (for $\zeta(3)$ specifically). – anon Feb 9 '12 at 5:21
• The same sum but for the second power ie. 1/(n^2-n))^2=(pi^2/3)-3 and 1/n^2=pi^2/6 – Vassilis Parassidis Feb 9 '12 at 5:41
• [1/n^2]-[1/(n^2-n)^2]=(3-pi^2/6) – Vassilis Parassidis Feb 9 '12 at 6:05
• @Vassilis: You are chronically forgetting the $\sum$ symbol. Why? At any rate, the observation in your last comment is of no help in finding $\zeta(3)$. – anon Feb 9 '12 at 6:28

I note that Bailey and Ferguson, Numerical results on relations between fundamental constants using a new algorithm, Mathematics of Computation 53 (1989) 649-656, proved that if there are integers $a,b,c,d,e$ such that $$a\zeta(3)=b\pi^3+c\pi^2+d\pi+e$$ then $$(a^2+b^2+c^2+d^2+e^2)^{1/2}\ge70,000,000,000,000$$ I'm sure that in the intervening 20-odd years even stronger results have been proved, but this is already enough to show that it is futile to search for a simple expression for $\zeta(3)$ in terms of rational numbers and small powers of $\pi$.

EDIT: It may also be of interest to note that there are many series which could fool someone into thinking they might be of use in relating $\zeta(3)$ to $\pi$. Here are two from Jolley's book, Summation of Series: \eqalign{(241)\qquad\sum_2^{\infty}{1\over(n^3-n)^2}&={\pi^2\over4}-{39\over16}\cr(341)\qquad\sum_0^{\infty}{(-1)^n\over(2n+1)^3}&={\pi^3\over32}}

• – anon Feb 9 '12 at 6:24
• There is a big connection between the zeta function and hyperelliptic equations. For n=5 we have 1/(n^2-n)^5=126-(pi^4/9)-(35pi^2/3). I am sorry for not using the proper symbols.I have calculated the function for several n. – Vassilis Parassidis Feb 9 '12 at 15:28