$\sum_{n=-\infty}^{\infty}\frac{x^n}{1-yq^n}=\sum_{n=-\infty}^{\infty}\frac{y^n}{1-xq^n}$? How to prove:$$\sum_{n=-\infty}^{\infty}\frac{x^n}{1-yq^n}=\sum_{n=-\infty}^{\infty}\frac{y^n}{1-xq^n}$$ where $|q|<|x|<1$ and $|q|<|y|<1$ . How to reach from left hand side to right hand side? However this identity shows that summation is summetric in $x,y.$ My effort is: $$\sum_{n=-\infty}^{\infty}\frac{x^n}{1-yq^n}=\sum_{n=-\infty}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{n=-\infty}^{-1}\sum_{k=0}^{\infty}x^ny^kq^{nk}+\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{n=1}^{\infty}\sum_{k=0}^{\infty}x^{-n}y^kq^{-nk}+\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^{-n-1}y^kq^{-nk-k}+\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{n=0}^{\infty}\sum_{k=1}^{\infty}x^{-n-1}y^{k-1}q^{-nk+n-k+1}+\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{n=0}^{\infty}\sum_{k=-\infty}^{-1}x^{-n-1}y^{-k-1}q^{nk+n+k+1}+\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x^ny^kq^{nk}\\=\sum_{k=-\infty}^{-1}\sum_{n=0}^{\infty}x^{-(n+1)}y^{-(k+1)}q^{(n+1)(k+1)}+\sum_{k=0}^{\infty}\frac{y^k}{1-xq^k}.$$  I don't know how to go further. Kindly help
 A: $$ 
\begin{align} 
\frac{1}{1-z} &= \sum_{k=\color{red}{0}}^{\infty}z^k = \sum_{k=\color{red}{1}}^{\infty}\frac{z^k}{z} \quad\colon\space |z|\lt1 \\[8mm] 
\sum_{m=0}^{\infty}\frac{x^m}{1-y\,q^m} &= \sum_{m=0}^{\infty}x^m\left[\sum_{n=0}^{\infty}\left(y\,q^m\right)^n\right] \\[2mm] 
&= \color{blue}{\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}x^m\,y^n\,q^{mn}} \\[2mm] 
&= \sum_{n=0}^{\infty}y^n\left[\sum_{m=0}^{\infty}\left(x\,q^n\right)^m\right] = \sum_{n=0}^{\infty}\frac{y^n}{1-x\,q^n} \\[8mm] 
\sum_{m=-\infty}^{-1}\frac{x^m}{1-y\,q^m} &= \sum_{m=-1}^{-\infty}\frac{x^m}{1-y\,q^m} = \sum_{m=1}^{\infty}\frac{x^{-m}}{1-y\,q^{-m}} \\[2mm] 
&= -\sum_{m=1}^{\infty}\frac{x^{-m}}{y\,q^{-m}}\,\frac{1}{1-y^{-1}\,q^m} = -\sum_{m=1}^{\infty}\frac{x^{-m}}{y\,q^{-m}}\,\left[\sum_{n=1}^{\infty}\frac{\left(y^{-1}\,q^m\right)^n}{y^{-1}\,q^m}\right] \\[2mm] 
&= \color{red}{-\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}x^{-m}\,y^{-n}\,q^{mn}} \\[2mm] 
&= -\sum_{n=1}^{\infty}\frac{y^{-n}}{x\,q^{-n}}\,\left[\sum_{m=1}^{\infty}\frac{\left(x^{-1}\,q^n\right)^m}{x^{-1}\,q^n}\right] = -\sum_{n=1}^{\infty}\frac{y^{-n}}{x\,q^{-n}}\,\frac{1}{1-x^{-1}\,q^n} \\[2mm] 
&= \sum_{n=1}^{\infty}\frac{y^{-n}}{1-x\,q^{-n}} = \sum_{n=-1}^{-\infty}\frac{y^n}{1-x\,q^n} = \sum_{n=-\infty}^{-1}\frac{y^n}{1-x\,q^n} 
\end{align} 
$$ 
Hence,
$$ 
\begin{align} 
\sum_{n=-\infty}^{\infty}\frac{x^n}{1-y\,q^n} &= \color{red}{\sum_{n=-\infty}^{-1}\frac{x^n}{1-y\,q^n}} + \color{blue}{\sum_{n=0}^{\infty}\frac{x^n}{1-y\,q^n}} \\[2mm] 
&= \color{red}{\sum_{n=-\infty}^{-1}\frac{y^n}{1-x\,q^n}} + \color{blue}{\sum_{n=0}^{\infty}\frac{y^n}{1-x\,q^n}} = \sum_{n=-\infty}^{\infty}\frac{y^n}{1-x\,q^n} 
\end{align} 
$$ 
Notice That: 
$$ |q|\lt|z|\lt1\implies \left|z\,q^k\right|\lt1\,\colon\,k\ge0 \quad\&\quad \left|z^{-1}\,q^k\right|\lt1\,\colon\,k\ge1 $$
