Introducing the multi-zeta function!

Contemplating:

\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}

My friend, and fellow math blogger, Owen, tried to tell me I was dealing with the multi-zeta function the other day, but the very general definition on WolframMathworld left me feeling a little mystified. I could see how the identities I was playing with fit in to it, but what was all of that other stuff (like \sigma_1,...,\sigma_k? I guess in my equation they are all 1? Why isn't it dealing with powers of the active variable in the product in the denominator? Why the extra level of subscripts?) So, I asked about it on stackexchange to gain more insight, and, thanks to Marni, I was promptly pointed to this lovely paper:

Now, I can see easily that the function is:

A(i_1, i_2, ... , i_k) = \sum_{n_1 > n_2 > \cdots > n_k > 1} \frac{1}{n_1^{i_1}n_2^{i_2} \cdots n_k^{i_k}}

With i_1, i_2, ... , i_k = 2.

More on this later.

Meaning one may repeat values,

Several series for powers of pi.

The Weierstrass factorization theorem gives us this identity:

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

Take the left side:

\left( 1 - \frac{1}{1}q^2 \right) \left( 1 - \frac{1}{4}q^2 \right) \left( 1 - \frac{1}{9}q^2 \right) \cdots

Thinking of this in a combinatorial way we can expand it. Each even power of q has a sum of combinations of the inverse squares as its coefficient. That is:

1 + \displaystyle \left(  q^2 \sum_{j_1=1}^{\infty} -j_1^{-2} \right) + \left( q^4 \sum_{j_1,j_2=1 \atop j_1 \neq j_2}^{\infty} (j_1j_2)^{-2} \right) + \left( q^6  \sum_{j_1,j_2,j_3=1 \atop j_i \neq j_k} (j_1j_2j_3)^{-2} \right) + \cdots + \displaystyle q^{2n} \sum_{j_1,...j_n=1 \atop j_i \neq j_k}^{\infty}( -1)^n (j_1j_2...j_n)^{-2} + \cdots

Now on the right side of
\displaystyle \prod_{n=1}^{\infty} \left( 1 -\frac{q^2}{n^2} \right) = \frac{\sin(\pi q)}{\pi q}

Expand using the Taylor series for the sine.

\frac{\sin(\pi q)}{\pi q} = 1 - \frac{(\pi q)^2}{3!} + \frac{(\pi q)^4}{5!} - \frac{(\pi q)^6}{7!} + \cdots

Equate the coefficients of the terms with the same exponent.:

  • 1  = 1
  • - \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

The second one is famous, more often written:

\frac{\pi ^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{25} \cdots .
And the rest of these series I do not know so well. Do you know them?