harmonic series - generating function

Question

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.)

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With the identity $\ds{\pars{1 - x}^{m} = \sum_{k = 0}^{\infty} {m \choose k}\pars{-x}^{k} = \sum_{k = 0}^{\infty}{k - m - 1 \choose k}x^{k}}$:

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\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}

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\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}

Answer 2

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*}