# Lambert series identity

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

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## 1 Answer

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.

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