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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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Your second one looks like a special case of Epstein zeta... –  Guess who it is. Jan 5 '12 at 16:35
...and the first one seems to be a special case of Barnes zeta. This might interest you... –  Guess who it is. Jan 5 '12 at 16:42
You can rewrite your first sum as $\sum_n \frac{{n-1}\choose{m-1}}{n^s}$. The numerator is a polynomial of degree $m-1$ in $n$, so I'm guessing it converges if and only if $s>m$. –  Thomas Andrews Jan 5 '12 at 16:44
@J.M. the second case isn't quite an Epstein zeta, since it includes non-positive $k_i$. It's clearly related, however. –  Thomas Andrews Jan 5 '12 at 17:04

1 Answer 1

up vote 5 down vote accepted

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 $q(n,m)<n^{\frac{m}2}$, 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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The asymptotics of your $q(n,m)$ are likely to be fairly complicated; it's close cousin to the sum-of-square function $r_m(n)$, and the value of that function is closely coupled to the divisors of $n$; asymptotic estimates of it AFAIK involve some fairly deep number theory. See mathworld.wolfram.com/SumofSquaresFunction.html for more details. –  Steven Stadnicki Jan 5 '12 at 17:25
@StevenStadnicki Yeah, I knew it would be a messy function, but it isn't quite $r_m(n)$, since that allows zero and counts squares of negatives and positives the same. Asymptotically, are they the same? –  Thomas Andrews Jan 5 '12 at 17:31
I'm pretty sure that asymptotically you have $r_m(n)-4q(n,m)\ll r_m(n)$, since the quantity on the left side is those sums that include zero and thus should be of smaller 'dimension' (specifically, I think it's just $r_{m-1}(n)$). –  Steven Stadnicki Jan 5 '12 at 17:37
Shouldn't that be $r_m(n)-2^mq(n,m)$? And shouldn't the result be bounded by $mr_{m-1}(n)$? –  Thomas Andrews Jan 5 '12 at 17:39
I would think that both series diverge for the same $s$, because the $l_1$ and $l_2$ norms are equivalent. –  Craig Feinstein Jan 5 '12 at 18:01

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