Can someone give me a short proof of the identity,
$$\sum\_{n=1}^\infty\frac{q^nx^{n^2}}{1-qx^{n}}+\sum\_{n=1}^\infty\frac{q^nx^{n(n+1)}}{1-x^n}=\sum\_{n=1}^\infty\frac{q^nx^n}{1-x^n}$$



Lambert series identity
It's just about using geometric series a lot. Indeed, we have


$\sum\limits\_{n=1}^\infty\frac{q^n x^{n^2}}{1-qx^n}=\sum\limits\_{n=1}^\infty\sum\limits\_{k=0}^\infty q^{n+k}x^{n(n+k)}=\sum\limits\_{m=1}^\infty\sum\limits\_{d\ge m}q^dx^{md}$


and


$\sum\limits\_{n=1}^\infty\frac{q^n x^{n(n+1)}}{1-x^n}=\sum\limits\_{n=1}^\infty\sum\limits\_{k=0}^\infty q^{n}x^{n(n+1+k)}=\sum\limits\_{m=1}^\infty\sum\limits\_{1\le d<m}q^dx^{md}$.


Therefore,


$\sum\limits\_{n=1}^\infty\frac{q^n x^{n^2}}{1-qx^n}+\sum\limits\_{n=1}^\infty\frac{q^n x^{n(n+1)}}{1-x^n}=
\sum\limits\_{m=1}^\infty\sum\limits\_{d\ge m}q^dx^{md}+\sum\limits\_{m=1}^\infty\sum\limits\_{1\le d<m}q^dx^{md}=
\sum\limits\_{m=1}^\infty\sum\limits\_{d=1}^\infty q^dx^{md}$.


However,


$\sum\limits\_{n=1}^\infty\frac{q^n x^{n}}{1-x^n}= \sum\limits\_{m=1}^\infty\sum\limits\_{d=1}^\infty q^dx^{md}$ as well, and we are done.


