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I am wondering if there is a way to evaluate or get a more useful expression for a sum of the following form: $$\sum_{i_d=1}^{\infty}\ldots\sum_{i_2=1}^{\infty}\sum_{i_1=1}^{\infty}x^{i_1\cdot i_2\cdots i_d},$$ where $|x|<1.$ For example, if $d=2$, then the sum is an example of a Lambert series and the exponents that appear are essentially given by (up to some index shuffling) this OEIS entry. In this case it is not difficult to obtain $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}x^{ij}=\sum_{j=1}^{\infty}\frac{x^{j}}{1-x^{j}},$$ which I can then evaluate for any $|x|<1.$

This can be viewed as summing over a $d$-dimensional non-negative integer lattice, so I've looked at things like "lattice sums" (tried to post a link here but not enough reputation points to post more than 2 links) but can't seem to find anything helpful there.

Even something for the case $d=3$ would be helpful.

guest6168834605
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Since you say you can evaluate the remaining infinite sum in the $d=2$ case, I gather that reducing the multiple sum to a single infinite sum would already be sufficient progress.

Then you want to count the number $\pi_d(n)$ of ways in which a positive integer $n$ can be written as an ordered product of positive integers. Find the prime factorisation, $n=\prod_ip_i^{k_i}$, and independently distribute $k_i$ balls over $d$ bins for each $i$, for a total of $\pi_d(n)=\prod_i\binom{k_i+d-1}{d-1}$ ways. Then your sum is

$$ \sum_{n=1}^\infty\pi_d(n)x^n\;, $$

the ordinary generating function for $\pi_d(n)$. Note that $\pi_d(n)$ is a multiplicative function, which allows you to calculate it without factorising.

joriki
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  • See also http://math.stackexchange.com/questions/1473925, which addresses the slightly different problem of counting the number of ways $n$ can be expressed as a product of any number of factors $\ge2$. – joriki Jun 24 '16 at 17:50
  • If you write $x = ,\mathrm{e}^{-\beta}$, the problem resembles a usual statistical physics problem. The usual difficulty is to find the 'number of configurations ' that makes $i_{1}\ldots i_{d} = n$ for a given $n$ which we call the number of states ( your $\pi_{d}\left(n\right)$ ) and it's given by $$ \sum_{i_{1} = 1}\ldots\sum_{i_{1} = 1}\delta_{n,\prod_{k = 1}^{d}i_{k}} $$

    It is usually the main task. Good luck.

     – Felix Marin Jun 24 '16 at 18:31
  • @FelixMarin: I don't understand. That's precisely the problem that my answer intended to solve. Are you saying that it doesn't? – joriki Jun 24 '16 at 18:41
  • You are quite right and that's the good direction to follow.

    I just want to highlight an analogy from Physics where it's a somehow current task and it's usually quite difficult. Fortunately, in Physics it's usual to look for the asymptotic behaviour ( 'for large $n$' ). It could be difficult too. I always follow your answers and they are always pretty fine and very professional.

    I'm sorry if I was misunderstood.

     – Felix Marin Jun 24 '16 at 18:49
  • @FelixMarin: I see. Thanks for the kind words, sorry about the misunderstanding. My first name is Felix, too, by the way :-) – joriki Jun 24 '16 at 18:57
  • @joriki This is what I thought, thanks. I suppose that there isn't much more simplifying that can be done. I know that the Dirichlet g.f. for $\pi_d(n)$ is $\zeta^d(s)$, but I can't seem to find any nice expressions for the ordinary generating function of $pi_d(n)$.... – guest6168834605 Jun 24 '16 at 21:19
  • @guest6168834605: If this is what you thought, it would have saved me a lot of time if you'd mentioned that in the question. – joriki Jun 24 '16 at 21:25