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For $n\in\mathbb{N}$ and $q\in(0,1)$, define $$f_{n}(q):=\sum_{i_{1},i_{2},\dots,i_{n}=1}^{\infty}\frac{q^{i_1+i_2+\dots+i_n}}{(1-q^{i_1+i_2})(1-q^{i_2+i_3})\dots(1-q^{i_{n-1}+i_n})(1-q^{i_n+i_1})}.$$

I would like to know something about the asymptotic behavior of $f_{n}(q)$, as $n\to\infty$.

Some remarks on possible ways to find an answer: I believe that having, for instance, a recurrence rule for $f_{n}(q)$ or a generating function would be helpful. So far, however, I wasn't able to find anything. Perhaps there is a relation between $f_{n}(q)$ and Jacobi theta functions (or some other special functions from the $q$-world) which could be helpful as well.

Maybe the expression for $f_{n}(q)$ is familiar to someone. I would appreciate any information concerning the sequence (even if it does not yield the asymptotic formula). Many thanks.

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    $\begingroup$ If you slightly modify your function as $$F_n(q,i):= \sum_{i_1,\dots,i_n} \frac{q^{i_1+\dots+i_n}}{(1-q^{i_1+i_2})\dots(1-q^{i_{n-1}+i_n})(1-q^{i_n+i})} \,,$$ then you get a recurrence: $$ F_n(q,i) = \sum_j \frac{q^j}{1-q^{j+i}}\, F_{n-1}(q,j) \,.$$ Hope this can help you. $\endgroup$
    – tituf
    Commented Apr 12, 2015 at 16:09

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Since $$ \frac{1}{1-q^{i_{1}+i_{2}}}\frac{1}{1-q^{i_{2}+i_{3}}}\ldots\frac{1}{1-q^{i_{n}+i_{1}}}>1, $$ we find \begin{align*} f_{n}(q) & >\sum_{i_{1}\ldots i_{n}\geq1}q^{i_1+\ldots+i_n}\\ & =\left(\sum_{i\geq1}q^{i}\right)^{n}\\ & =\left(\frac{1}{1-q}\right)^{n}\to\infty. \end{align*} Additionally, When $q<\frac{1}{2}, $$\frac{\sqrt{q}}{1-q^{2}}<1,$ and for all $n\geq1$ \begin{align*} \left(1-q^{2}\right)^{n}+\left(q^{2}\right)^{n} & \leq1\\ \Leftrightarrow\left(1-q^{2}\right)^{n} & \leq\left(1-q^{2n}\right)\\ \Leftrightarrow q^{n}\left(1-q^{2}\right)^{n} & \leq q^{n}\left(1-q^{2n}\right)\\ \Leftrightarrow\frac{q^{n}}{1-q^{2n}} & \leq\left(\frac{q}{1-q^{2}}\right)^{n} \end{align*} So \begin{align*} f_{n}(q)=\sum_{i_{1}\ldots i_{n}\geq1}\frac{\left(\sqrt{q}\right)^{i_{1}+i_{2}}}{1-q^{i_{1}+i_{2}}}\frac{\left(\sqrt{q}\right)^{i_{2}+i_{3}}}{1-q^{i_{2}+i_{3}}}\ldots\frac{\left(\sqrt{q}\right)^{i_{n}+i_{1}}}{1-q^{i_{n}+i_{1}}} & \leq\sum_{i_{1}\ldots i_{n}\geq1}\left(\frac{\sqrt{q}}{1-q^{2}}\right)^{2\left(i_{i}+\ldots+i_{n}\right)}\\ & =\left(\sum_{i=1}^{\infty}\left(\frac{q}{\left(1-q^{2}\right)^{2}}\right)^{i}\right)^{n}\\ & =\left(\frac{\left(1-q^{2}\right)^{2}}{\left(1-q^{2}\right)^{2}-q}\right)^{n} \end{align*} Therefore, when $q\in(0,\frac 12)$ $$\left(\frac{1}{1-q}\right)^{n}\leq f_{n}(q)\leq\left(\frac{\left(1-q^{2}\right)^{2}}{\left(1-q^{2}\right)^{2}-q}\right)^{n}.$$

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