2
$\begingroup$

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.

  • 0
    Your second one looks like a special case of [Epstein zeta](http://mathworld.wolfram.com/EpsteinZetaFunction.html)...2012-01-05
  • 0
    ...and the first one seems to be a special case of [Barnes zeta](https://en.wikipedia.org/wiki/Barnes_zeta_function). [This](http://www.ieec.fcr.es/espanyol/recerca/ftc/eli/jpa22ee931_89rev.pdf) might interest you...2012-01-05
  • 0
    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 $s>m$.2012-01-05
  • 0
    @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

5

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 $s>m$.

  • 1
    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 http://mathworld.wolfram.com/SumofSquaresFunction.html for more details.2012-01-05
  • 0
    @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?2012-01-05
  • 0
    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 $r_{m-1}(n)$).2012-01-05
  • 0
    Shouldn't that be $r_m(n)-2^mq(n,m)$? And shouldn't the result be bounded by $mr_{m-1}(n)$?2012-01-05
  • 0
    On the other hand, asymptotics for $r_m$ may not be so bad, since you can treat it as essentially counting the 'shell area' for an $m$-dimensional sphere; this gives you easy estimates for $\Sigma_{i=0}^n r_m(i)$ as $C_mn^{m/2}$ for some $C_m$ (dependent on $m$, of course), but using that to get estimates on the weighted sums may be harder. On the other hand, that may be enough to give estimates up to polylog($n$) factors, which would be enough to establish the radius of convergence ($r_m$ is 'roughly' $n^{m/2-1}$, so $q(n,m)$ is 'roughly' $2^{-m}n^{m/2-1}$.)2012-01-05
  • 0
    Ahh, yes - I was thinking of low-dimensional cases and got my factors slightly tangled up. Yes, it's $r_m(n)-2^mq(n,m)$, and your bound is also right; as I said, that should be of essentially 'lower order' and so asymptotically moot. (Although that changes my radius of convergence too, so mea culpa there.)2012-01-05
  • 2
    I would think that both series diverge for the same $s$, because the $l_1$ and $l_2$ norms are equivalent.2012-01-05
  • 0
    @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