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

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.

share|cite|improve this question
Your second one looks like a special case of Epstein zeta... – J. M. Jan 5 '12 at 16:35
...and the first one seems to be a special case of Barnes zeta. This might interest you... – J. M. 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
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 $s>m$.

share|cite|improve this answer
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 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

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.