Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Playing around with elementary symmetric functions I tried to generalize that to infinite series and arrived at the well known concept of MZV ("multiple zeta values"). At the moment I'm only interested on that MZV, where all exponents are equal, and I compute that for the convergent cases by the rules of composition of the elementary symmetric functions by the according powersums and vice versa.
Meanwhile I've found a lot of material on MZV via google (mostly much "high-tech"), but very little on the simpler question of MZV with equal exponents (might be too introductory and there might not be too much interesting in it). Then I did not find anything about some asymptotic expression for the $\small \zeta(s),\zeta(s,s),\zeta(s,s,s),\ldots$ when $\small s=1+ \delta, \delta \to 0 $ except the standard remark that they become singular.
Meanwhile I've tried to find some expression depending on the $\small \delta $ by numerical approximation using Pari/GP and also by the expression of the $\small \zeta(s) $ by a power series (and then develop the symbolic expression of the MZV's in that terms of power series), but do not yet have an conclusive general formulation.

Is there a simple answer for this or else some article around, which deals explicitely with that question?

[update after anon's comment] While the sequence of $\small \zeta(2),\zeta(2,2),\zeta(2,2,2),\ldots $ can be done nicely with the formulae for the elementary symmetric polynomials and using the convergent series $\small \zeta(1\cdot 2),\zeta(2\cdot 2),\zeta(3\cdot 2),\ldots$ at the place of the power-sums, clearly $\small \zeta(1),\zeta(1,1),\zeta(1,1,1) $ cannot be done although all except the first zetas used for this in the elementary-symmetric-polynomials formula: $\small \zeta(1 \cdot 1),\zeta(2 \cdot 1),\zeta(3 \cdot 1), \ldots $ are nonsingular. When I follow the idea of Ramanujan in his concept of zeta-regularization/-summation and set for the singularity $\zeta(1)$ the Euler-Mascheroni-constant $\small \gamma_0$ as first entry:$\small \gamma_0,\zeta(2 \cdot 1),\zeta(3 \cdot 1), \ldots $ at the place of the power-sums and determine the variants $\small \zeta^*(1),\zeta^*(1,1),\zeta^*(1,1,1),\ldots $ then the sum of all that MZV's adds just to 0 . ( If I use $\small \gamma_0-1,\zeta(2)-1,\zeta(3)-1, \ldots $ the MZV's sum up to $\small -\frac12 $, not much surprising) Just a nice finding meanwhile ...

share|cite|improve this question
We can write $e_n=P(p_1,\cdots,p_n)$ in elementary symmetric and power polynomials hence $$\zeta(\underbrace{s,\cdots,s}_n)=P\big(\zeta(s),\cdots,\zeta(ns)\big).$$ To get the asymptotic expansion of this up to $o(1)$, we would need to compute the Laurent expansion of $\zeta(s)$ up to around $(s-1)^{+n}$. Not to mention I don't think $P$ has a nice form to it. I doubt there's a simple answer to this. – anon Apr 28 '12 at 6:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.