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What is this series called (if it has a name)? When does it diverge without analytic continuation and when does it diverge with analytic continuation?

$\sum_{k_1,\dots,k_m=1}^{\infty} (k_1+\dots+k_m)^{-s}$, where $\Re{(s)}>0$.

What about this series?

$\sum_{k_1,\dots,k_m=1}^{\infty} (k_1^2+\dots+k_m^2)^{-s/2}$, where $\Re{(s)}>0$.

I looked up multi-dimensional zeta function, but couldn't find anything.

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    @J.M. the second case isn't quite an Epstein zeta, since it includes non-positive $k_i$. It's clearly related, however.2012-01-05

1 Answers 1

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You can rewrite your first sum as $\sum_n \frac{{n-1}\choose{m-1}}{n^s}$, because ${n-1}\choose{m-1}$ is the number of ways of writing $n$ as the sum of $m$ positive integers.

Since ${n-1}\choose{m-1}$ is a polynomial of degree $m-1$ in $n$, this series converges only when $\sum_n {n^{m-1-s}}$ converges, which is precisely when $s>m$.

Letting $q(n,m)$ be the number of ways of writing $n$ as the sum of $m$ positive squares, the second sum is $\sum_n \frac{q(n,m)}{n^{s/2}}$. So you'll need some estimate/bounds for $q(n,m)$ to figure out the values of $s$ for which this converges.

It's pretty easy to see that $q(n,m), for example, which shows convergence if $\frac{s}2>\frac{m}2 + 1$, but it seems likely that you'd have convergence for smaller $s$.

By coment below, since $\frac{1}{ m}(k_1+...+k_m)^2\leq k_1^2+...+k_m^2\leq (k_1+...+k_m)^2$, we see that if one of these series converges, then the other must, so the second series likewise converges exactly when $s>m$.

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    @CraigFeinstein - Ah, yes, returning to the original form of the equations, we can see that the first converges if and only if the second converges. Added to my answer.2012-01-05