I've run into:

$$\sum_{x=1}^{\infty} {x^a\over 1-q^{x}}, \ s.t.\ q\in \mathbb N>1 \ or \ q\in (0, 1),\ a \in \mathbb N$$

I am interested mostly in the cases where $a = 1$ or $ a = 2$

Things I've done so far: reference a few places on basic hypergeometric series, not limited to something that looks remotely like what I'm interested in: Take a look at the "Simple Series" section, the first example

I'm looking for formulae that are "short and simple", ideally. Of course, if that cannot be done, I'll settle for computationally efficient with a static number of terms regardless of input.

EDIT: I've made some progress, please see: JJacquelin's answer to this question, can someone help get their attention (I cannot comment yet). If we cannot, should not, or do not wish to get their attention through contact, perhaps explaining some of the manipulations in their answer might help us here. For example, pulling out the $1 \over m$ from the sum in the second to last line to obtain the integral shown, why was that done? What technique would apply if it were $1 \over m!$ instead?

Secondly, I've found that theta functions may be involved somehow:See GEdgar's answer here as well as Paramanand Singh's answer here.

  • $\begingroup$ edited for domain clarifications, $q$ can indeed be a rational between $0$ and $1$, or an integer $>\ 1$. $\endgroup$ Jan 24, 2019 at 5:23
  • $\begingroup$ Would the person who downvoted me care to let me know why I've received a downvote? I would love to edit the question if I'm in violation of the (extensive) rules. $\endgroup$ Jan 24, 2019 at 17:09
  • $\begingroup$ What do you mean by "closed form"? Usually, a finite sum is considered to be a very satisfactory state of affairs; I can't quite imagine getting more "closed" than that. Anyway, have you looked at Lambert series? $\endgroup$
    – Linas
    Jan 24, 2019 at 18:22
  • $\begingroup$ @Linas I'd like a partial sum formula for the above, ideally. $\endgroup$ Jan 25, 2019 at 4:55
  • 1
    $\begingroup$ @DavidRoberts The intention was not to trigger you, I apologize. Can you (please) suggest the proper notation to express what is already intended to be expressed? $\endgroup$ Jan 25, 2019 at 6:38

2 Answers 2


Please note that, if I am not mistaken, for the case of $q \in (0,1)$ the sum diverges, since

$$ \sum_{x=1}^{\infty} {x^a\over 1-q^{x}} = \sum_{x=1}^{\infty} \sum_{l=0}^{\infty}{x^a q^{l x}} > \sum_{x=1}^{\infty} x^a = \infty $$

About your question concerning JJacquelin's manipulations, the reason he integrates is because he wants to get rid of the term $1/m$ in the denominator that prevents him from using his previous result. Doing this integration yields

$$ \int \frac{1}{x} \sum_{m=1}^\infty \frac{(x y)^m}{1-y^m} dx = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int x^{m-1}dx= \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m} = \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m} $$

If it had $m!$ in the denominator, as you ask, the same manipulation would have led to multiple integrations

$$ \sum_{m=1}^\infty \frac{(x y)^m}{(1-y^m)m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \frac{x^m}{m!} = \sum_{m=1}^\infty \frac{y^m}{1-y^m} \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \sum_{m=1}^\infty \frac{y^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\sum_{m=1}^\infty \frac{(x_1 y)^m}{1-y^m}dx_1 dx_2 \dots dx_m =\\ \int_0^{x} \cdots \int_0^{x_3} \int_0^{x_2} \frac{1}{x_1^m}\left [ \psi_y \left( 1 + \frac{\ln(x_1)}{\ln(y)} + \ln(1-y) \right) \right]dx_1 dx_2 \dots dx_m $$

  • $\begingroup$ Note @AmirSagiv that $x$ is not a variable but the iterator. That is $\sum_{x=1}^{\infty} x^a = 1^a + 2^a + 3^a$ etc. $\endgroup$
    – Sotiris
    Jan 26, 2019 at 19:42
  • $\begingroup$ Careful about the domain given $Re(a)$. So where you start for analytic continuation. $\endgroup$
    – AHusain
    Jan 26, 2019 at 21:06
  • $\begingroup$ Well, Jjacqueline tried to find a closed form for the expression he was studying and the closest he manage to go was near the q-digamma function being one integral "away". In this sense, you are correct to say that using multiple integrals places us even "farther" from a closed form, but this is the result his manipulations would yield. Unfortunately, I do not know of a simpler expression for this. $\endgroup$
    – Sotiris
    Jan 26, 2019 at 23:44
  • $\begingroup$ Please note also that the simple series expressions you found in Wikipedia, are for the expression $\sum_{x=1}^{\infty}\frac{a^x}{1-q^{x}}$, using your symbols. $\endgroup$
    – Sotiris
    Jan 26, 2019 at 23:46

Not an answer, but a research idea: first, replace $x$ by $k$ then write your sum as follows:

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{k=1}^\infty \left(c_k \frac{k^a}{q^k}\right) \frac{q^k}{1-q^k}$

where $c_k=0$ for $k>n$ and 1 otherwise. This is a Lambert series; see wikipedia. That is, let

$a_k = \left(c_k \frac{k^a}{q^k}\right)$

and then apply a Dirichlet convolution (see wikipedia) to get

$b_m = (a*1)(m) = \sum_{k|m} a_k$

and thus

$\sum_{k=1}^n \frac{k^a}{1-q^k} = \sum_{m=1}^\infty b_m q^m$

So now you have an infinite sum, in place of a finite one. A different, less fancy way to say what I just said is to just write

$\frac{1}{1-q^k} = \sum_{m=0}^\infty (q^k)^m$

and just plug that in, and then exchange order of summation. That way, you don't have a pesky denominator; you traded it for an infinite sum, which is pesky in a different way.

I suspect my suggestion above is completely useless; I cant quite figure out what you want. You might also find more willing help on math exchange instead of mathoverflow?

  • $\begingroup$ I've edited the question to reflect your point about extracting a partial sum formula from something that is itself already a discrete sum. It now makes a bit more sense, the goal: to extract the partial sum formula from the infinite sum. $\endgroup$ Jan 26, 2019 at 17:12
  • $\begingroup$ We also care that the computational efficiency of the partial sum (whichever we find). We'd like something with a bounded number of terms not related to the size of the input in any way. $\endgroup$ Jan 26, 2019 at 17:25

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