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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?

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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.

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  • $\begingroup$ @DavidH: you are right, now fixed. $\endgroup$ – Jack D'Aurizio Dec 11 '14 at 20:23
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    $\begingroup$ We're now in agreement, so previous comment removed and +1 added. :) $\endgroup$ – David H Dec 11 '14 at 20:31
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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 $z,x\notin[1,\infty)$,

$$\begin{align} \sum_{n=1}^{\infty}H_{n}z^{n} &=-\frac{\ln{\left(1-z\right)}}{1-z}\\ \implies \sum_{n=1}^{\infty}H_{n}z^{n-1}&=-\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\\ \implies \int_{0}^{x}\sum_{n=1}^{\infty}H_{n}z^{n-1}\,\mathrm{d}z&=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\,\mathrm{d}z\\ \implies \sum_{n=1}^{\infty}H_{n}\frac{x^{n}}{n}&=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z\left(1-z\right)}\,\mathrm{d}z\\ &=-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{z}\,\mathrm{d}z-\int_{0}^{x}\frac{\ln{\left(1-z\right)}}{1-z}\,\mathrm{d}z\\ &=\operatorname{Li}_{2}{\left(x\right)}+\frac12\ln^2{\left(1-x\right)}\\ &=-\operatorname{Li}_{2}{\left(\frac{x}{x-1}\right)}.\\ \end{align}$$

Note, the simplification made in the last step made using Landen's dilogarithm identity.

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