harmonic series - generating function I am currently learning about generating functions and
I found an interesting one for harmonic series, $\dfrac{\log(1-x)}{x-1}$.
Is there any hope I could get a formula for $n$th coefficient out of this? The $n$th derivative looks messy...
In absence of formula, can I at least get some asymptotic information, like that harmonic series diverges? (Can be shown more simply, I know.)
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With the identity
$\ds{\pars{1 - x}^{m} = \sum_{k = 0}^{\infty}
{m \choose k}\pars{-x}^{k} =
\sum_{k = 0}^{\infty}{k - m - 1 \choose k}x^{k}}$:

\begin{equation}
\begin{array}{l}
\mbox{Derivative respect of}\ \ds{m}:
\\\ 
\ds{\pars{1 - x}^{m}\ln\pars{1 - x} = 
\sum_{k = 0}^{\infty}\bracks{\partiald{}{m}{k - m - 1 \choose k}}x^{k}}
\\[5mm]
\mbox{The limit}\ds{\ m \to - 1}:
\\
\ds{-\,{\ln\pars{1 - x} \over 1 - x} =
\sum_{k = 0}^{\infty}\color{#f00}{\bracks{-\,\partiald{}{m}{k - m - 1 \choose k}}
_{\ m\ =\ - 1}}\ x^{k}}
\end{array}
\label{1}\tag{1}
\end{equation}

\begin{align}
&\color{#f00}{\bracks{-\,\partiald{}{m}{k - m - 1 \choose k}}_{\ m\ =\ - 1}} =
\left.\vphantom{\Huge A}-\,\partiald{}{m}\bracks{\Gamma\pars{k - m} \over k!\,\Gamma\pars{-m}}\right\vert_{\ m\ =\ - 1}
\\[5mm] = &
-\,{1 \over k!}\,
{-\Gamma\, '\pars{k + 1}\Gamma\pars{1} + \Gamma\, '\pars{1}\Gamma\pars{k + 1}\over \Gamma^{2}\pars{1}}
\\[5mm] = &\
-\,{1 \over k!}\bracks{%
{-\Gamma\pars{k + 1}\Psi\pars{k + 1} +
\Gamma\pars{1}\Psi\pars{1}\Gamma\pars{k + 1}}}
\\[5mm] = &\
\Psi\pars{k + 1} - \Psi\pars{1} = \color{#f00}{H_{k}}
\\[1cm]
\stackrel{\mbox{see}\ \eqref{1}}{\implies} &\ \,\,\,
\bbox[10px,#ffe,border:1px dotted navy]{\ds{%
-\,{\ln\pars{1 - x} \over 1 - x} = \sum_{k = 1}^{\infty}H_{k}\, x^{k}}}
\end{align}
A: It is useful to notice that multiplication of a series $A(x)=\sum_{n=0}^\infty a_nx^n$ with the geometric series $\frac{1}{1-x}$ transforms the sequence  $(a_n)_{n\geq 0}$ to a  sequence of  sums $\left(\sum_{k=0}^na_k\right)_{n\geq  0}$. We obtain
\begin{align*}
\frac{1}{1-x}A(x)&=\frac{1}{1-x}\sum_{n=0}^\infty a_n x^n\\
&=\sum_{n=0}^\infty \left(\sum_{k=0}^n a_k\right) x^n\tag{1}
\end{align*}
So, it is sufficient to determine the $n$-th coefficient of $A(x)$ in order to also know the $n$-th coefficient of $\frac{1}{1-x}A(x)$.

Since the series expansion of $-\log (1-x)$ is known to be
  \begin{align*}
-\log(1-x)=\sum_{n=1}^\infty \frac{x^n}{n}\qquad\qquad\qquad |x|<1
\end{align*}
  it follows from (1)
  \begin{align*}
-\frac{\log(1-x)}{1-x}&=\frac{1}{1-x}\sum_{n=1}^\infty \frac{x^n}{n}
=\sum_{n=1}^\infty \left(\sum_{k=1}^n\frac{1}{k}\right)x^n\\
&=\sum_{n=1}^\infty H_nx^n
\end{align*}

