# An infinite series involving Jordan's totient function

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 and $$A_k(x)$$ is the kth Eulerian polynomial.

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.

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.