Can you provide a proof for the following claim:



> 
> $$-\displaystyle\sum\_{n=1}^{\infty}\frac{J\_k(n)}{n} \cdot \ln\left(1-x^n\right)=\frac{x \cdot A\_{k-1}(x)}{(1-x)^k} \quad \text{for} \quad |x| < 1 \quad \text{and} \quad k>1$$
> where $J\_k(n)$ is the [Jordan's totient function](https://en.wikipedia.org/wiki/Jordan%27s_totient_function) and $A\_k(x)$ is the kth [Eulerian polynomial](https://en.wikipedia.org/wiki/Eulerian_number).
> 
> 
> 


The first few Eulerian polynomials are: $A\_1(x)=1 , A\_2(x)=x+1 , A\_3(x)=x^2+4x+1$ , etc.


The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJw9jEEKgzAQRfc5xSxnQqKmLtNcoIcQFNMSEkcxjQild29KoZ-3-fD-L5MzXY0Vp9Nd0_dWRHex4oZRMblXuCPw1SjoFOSyzOFAVjBXhggSltVPoWRkaGEmIvsW2VUP2RlVJvW7aVmm9YFGnwNXR2x74CfqLE3zb6f0x5jQl-T3MPK2JozaELXfGQ2R7AfRYDEd&lang=gp&interacts=eJyLjgUAARUAuQ==).



An infinite series involving Jordan's totient function
The right hand side evaluates to
$$\sum\_{m=0}^{\infty} m^{k-1} x^m,$$
and so it remains to verify that the coefficient of $x^m$ in the l.h.s. is also $m^{k-1}$.


By differentiating the l.h.s., we have
\begin{split}
[x^m]\ \bigg(-\sum\_{n\geq 1} \frac{J\_k(n)}{n} \ln(1-x^n)\bigg) &= \frac1m [x^{m-1}]\ \sum\_{n\geq 1} J\_k(n)\frac{x^{n-1}}{1-x^n}\\
&=\frac1m \sum\_{n|m} J\_k(n) \\
&= m^{k-1}
\end{split}
as required.


