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In the same vein as:

$ \frac{\pi ^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{25} \cdots $

Starting with:

$ \displaystyle \prod_{n=1}^{\infty} \left( 1 -\frac{q^2}{n^2} \right) = \frac{\sin(\pi q)}{\pi q}$

I've noticed that:

$ - \frac{\pi ^2}{3!} = \displaystyle \sum_{j_1=1}^{\infty} -j_1^{-2} $

$ \frac{\pi ^4}{5!} = \displaystyle \sum_{j_1,j_2=1 \atop j_1 \neq j_2}^{\infty} (j_1j_2)^{-2}$

$ - \frac{\pi ^6}{7!} = \displaystyle \sum_{j_1,j_2,j_3=1 \atop j_i \neq j_k} - (j_1j_2j_3)^{-2}$

$ \vdots $

$ \frac{\pi ^{2n}}{(2n+1)!} = \displaystyle \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty} (j_1j_2...j_n)^{-2}$

$ \vdots $

(Steps shown here: )

Is there a more direct way to reach the same result that avoids a high power theorem like the Weierstrass factorization theorem ... which is what I use.

I'm enjoying playing with these concepts so I'd also like reading recommendations.

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The following thread contains multiple solutions to the general problem, and may be of interest:… – Eric Naslund Apr 1 '11 at 22:25
That's a great thread, I wonder how many of them can be modified to produce the other less elegant identities above. – futurebird Apr 1 '11 at 22:28
I just changed the title to make it look better and it got bumped to the first page... :( – futurebird Apr 6 '11 at 22:43
up vote 9 down vote accepted

This is an example of a Multiple Zeta Value, namely $\zeta (2,2,2,\cdots,2)$. On the page

there are several relations satisfied by such MZVs. For example,

$\zeta (2,2,2,2, \cdots) = (2n + 1) \zeta (3,1,3,1,\cdots)$

where there are $2n$ copies of $2$ on the left, and $n$ blocks of $(3,1)$ on the right. So for $n = 1$ we have $\zeta (3,1) = 2 \pi^4 / 6!$. A good reference for this is

which says in general that

$\zeta (\{ 3,1 \}^{n}) = \frac{2 \pi^{4n}}{(4n + 2)!}$

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The original paper of relevance, containing the proof you require, is here:… – Kea Apr 1 '11 at 23:24

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