Series involving Euler totient function $\varphi(n)$

Question

Prove that:

$$\sum_{n=1}^\infty \frac{x^n\varphi(n)}{1-x^n}=\frac{x}{(1-x)^2}$$

where $\varphi$ is the Euler totient function.

I can see in Wolfram Alpha that this is the case, but how to prove it?

Answer 1

Expanding out $\frac{1}{1-x^n}$ in a power series leads to $$ \sum_{n=1}^{\infty}\frac{x^n\phi(n)}{1-x^n}=\sum_{n=1}^{\infty}\phi(n)x^n\sum_{k=0}^{\infty}x^{nk}=\sum_{n=1}^{\infty}\phi(n)\sum_{k=1}^{\infty}x^{nk} $$ For a given $m\geq 1$, the coefficient of $x^m$ in the above sum is $\sum_{d|m}\phi(d)=m$, hence $$ \sum_{n=1}^{\infty}\frac{x^n\phi(n)}{1-x^n}=\sum_{m=1}^{\infty}mx^m$$

Finally, differentiating the geometric series $$ \frac{1}{1-x}=\sum_{m=0}^{\infty}x^m$$ and multiplying by $x$ shows that $$\sum_{m=1}^{\infty}mx^m=\frac{x}{(1-x)^2}$$