# generating function for harmonic sequence

How I can find the generating function for the sequence $$\frac{ H_n } n$$ where $H_n$ -harmonic numbers. I know that $$\sum\limits_{k=1}^{\infty}H_kz^k = -\frac{\ln(1 - z)}{1 - z}$$ So, what should I do? Just find generating function for the sequence $$\sum\limits_{k=1}^{\infty}H_kz^k\frac{1}{k}$$ or what?

Integrate both sides of your identity between $0$ and $t$. You'll get: $$\frac{1}{2}\log^2(1-t)=\sum_{k=1}^{+\infty}\frac{H_k}{k+1} t^{k+1} = \sum_{k\geq 2}\frac{H_{k-1}}{k}t^k.$$ Since $H_{k-1}=H_k-\frac{1}{k}$, by adding $\operatorname{Li}_2(t)$ we are done.
Starting from the generating function for the harmonic numbers, a generating function for the sequence $\frac{H_{n}}{n}$ may be found as follows: for $z,x\notin[1,\infty)$,
\begin{align} \sum_{n=1}^{\infty}H_{n}z^{n} &=-\frac{\ln{\left(1-z\right)}}{1-z}\\ \implies \sum_{n=1}^{\infty}H_{n}z^{n-1}&=-\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\\ \implies \int_{0}^{x}\sum_{n=1}^{\infty}H_{n}z^{n-1}\,\mathrm{d}z&=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\,\mathrm{d}z\\ \implies \sum_{n=1}^{\infty}H_{n}\frac{x^{n}}{n}&=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\,\mathrm{d}z\\ &=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z}\,\mathrm{d}z-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{1-z}\,\mathrm{d}z\\ &=\operatorname{Li}_{2}{\left(x\right)}+\frac12\ln^2{\left(1-x\right)}\\ &=-\operatorname{Li}_{2}{\left(\frac{x}{x-1}\right)}.\\ \end{align}