Asymptotic Approximation of $\sum_{n_{2i-1} \ne n_{2i}}\sqrt{n_1\cdots n_k} x^{n_1 + \cdots +n_k}$

Question

For $1\le n_i \lt \infty$ and $|x|\lt 1$, let $$S_k=\sum_{n_{2i-1} \ne n_{2i}}\sqrt{n_1\cdots n_k} x^{n_1 + \cdots +n_k}$$ I would like to find a simpler expression that is asymptotically equivalent to $S_k$ as $k\to\infty$. Since $S_k$ is a $k-$dimensional sum and has that added condition, I’m out of any good ideas. The only thought I’ve had is of squaring $S_k$, to get $$S^2_{k} = \sum_{n_{2i-1}\ne n_{2i}}n_1 \cdots n_{k} (x^2)^{n_1 +\cdots +n_k} +2S_{2k} $$ which doesn’t seem helpful. How to accomplish this task, or atleast simplify $S_k$?

Answer 1

First note that, in the absence of the constraint that $n_{2i}\neq n_{2i-1}$ for all $i$, your $k$-dimensional sum factors into $k$ identical one-dimensional sums. In other words, it's separable. The result is the $k$-th power of the single sum, which can be written in terms of the polylogarithm: $$ \sum_{n_1,n_2,\ldots n_k=1}^{\infty}\sqrt{n_1 n_2\ldots n_k}x^{n_1+n_2+\ldots+n_k}=\sum_{n_1=1}^{\infty}\sqrt{n_1}x^{n_1}\sum_{n_2=1}^{\infty}\sqrt{n_2}x^{n_2}\cdots \sum_{n_k=1}^{\infty}\sqrt{n_k}x^{n_k}=\left(\sum_{m=1}^{\infty}\sqrt{m}x^{m}\right)^{k}=\left({\text{Li}}_{-1/2}(x)\right)^k. $$ Now, the constraint couples every other $n_i$ to the previous one. So in the presence of the constraint, assuming $k$ is even, the sum factors into $k/2$ identical two-dimensional sums: $$ {\sum_{n_1,n_2,\ldots n_k=1}^{\infty}}^{\prime}\sqrt{n_1 n_2\ldots n_k}x^{n_1+n_2+\ldots+n_k}=\sum_{n_1\neq n_2=1}^{\infty}\sqrt{n_1 n_2}x^{n_1 + n_2}\cdots \sum_{n_{k-1}\neq n_k=1}^{\infty}\sqrt{n_{k-1} n_k}x^{n_{k-1}+n_k}=\left(\sum_{l\neq m=1}^{\infty}\sqrt{l m}x^{l + m}\right)^{k/2}. $$ The two-dimensional sum is just $$ \sum_{l\neq m=1}^{\infty}\sqrt{l m}x^{l + m}=\sum_{l=1}^{\infty}\sum_{m=1}^{\infty}\sqrt{lm}x^{l+m}-\sum_{l=1}^{\infty}lx^{2l}=\left({\text{Li}}_{-1/2}(x)\right)^2-{\text{Li}}^{-1}(x^2) =\left({\text{Li}}_{-1/2}(x)\right)^2-\frac{x^2}{(1-x^2)^2}. $$ And so for even $k$, the sum is exactly $$ \left(\left({\text{Li}}_{-1/2}(x)\right)^2-\frac{x^2}{(1-x^2)^2}\right)^{k/2}. $$ For odd $k$, you have one additional unconstrained $n_i$ being summed over, so the result in that case is exactly $$ \left(\left({\text{Li}}_{-1/2}(x)\right)^2-\frac{x^2}{(1-x^2)^2}\right)^{\lfloor{k/2}\rfloor} {\text{Li}}_{-1/2}(x) . $$