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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}.$$