# Generating function of square harmonic numbers

During my study of generating functions, I was able to calculate the generating function of the sequence of harmonic numbers $H_n$: $$\sum_{n=1}^\infty H_nx^n=\frac{\ln(1-x)}{x-1}$$ However, I also tried to find generating functions for $H_n^2$ and $H_n^3$, with which I was unsuccessful (the rearrangement method I used for the generating function of $H_n$ didn't reduce as nicely for $H_n^2$ and $H_n^3$).

Any hints about how to find $$\sum_{n=1}^\infty H_n^2x^n=\space ?$$ and $$\sum_{n=1}^\infty H_n^3x^n=\space ?$$ Please don't write a full answer and spoil it for me - I just want a hint.

• The form for the cubes generating function doesn't look particularly nice. See here from another post. – Foobaz John Nov 22 '17 at 0:48
• @FoobazJohn Ooh, yuck. – Frpzzd Nov 22 '17 at 0:50
• $f(z) = \frac{\log(1-z)}{z-1}$, then $\sum_n H_n^2 z^n = \frac{1}{2i\pi}\int_{|s| = 1/2} f(s) f(z/s) \frac{ds}{s}$ which should reduce to something with $\text{Li}_2(z)$ by partial fraction and integration by parts – reuns Nov 22 '17 at 1:22
• @Nilknarf Hint: try to exploit the results of my recent general solution to math.stackexchange.com/questions/1073628/… – Dr. Wolfgang Hintze Nov 22 '17 at 11:50
• First GF is possibly $\sum_{n=1}^\infty H_n^2x^n=\frac{1}{1-x}Li_2(x)$ – James Arathoon Nov 22 '17 at 12:27

We have that $$f(x)=\sum_{n\geq 0}a_n x^n\quad\Longleftrightarrow\quad \frac{f(x)}{1-x}=\sum_{n\geq 0}A_n x^n$$ where $A_n = a_0+a_1+\ldots+a_n$. In order to find the OGF of $H_{n}^2$ it is enough to find the OGF of $$H_{n+1}^2-H_{n}^2 = \left(H_{n+1}-H_n\right)\left(H_{n+1}+H_n\right)=\frac{2H_n}{n+1}+\frac{1}{(n+1)^2},$$ and while the OGF of $\frac{1}{(n+1)^2}$ is clearly related with $\text{Li}_2(x)=\sum_{n\geq 1}\frac{x^n}{n^2}$, the OGF of $\frac{H_n}{n+1}$ can be deduced by applying termwise integration to the OGF of $H_n$. It follows that $$\sum_{n\geq 1}H_n^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x}\tag{A}$$ and the (more involved) OGF of $H_n^3$ can be computed through the same trick. It is useful to consider that the Taylor series of $\log(1-x)^k$ is related to Stirling numbers of the first kind.

• Ah-ha, so that's the trick. Well done, and thanks again! – Frpzzd Nov 23 '17 at 0:11


Hereafter, $\ds{\bracks{\cdots}}$ is an Iverson Bracket. Namely, $\ds{\bracks{P} = 1}$ whenever $\ds{P\ \mbox{is}\ \color{red}{\texttt{true}}}$ and $\ds{0\ \color{red}{\mbox{otherwise}}}$.

\begin{align} \left.\sum_{n = 1}^{\infty}H_{n}^{2}x^{n}\,\right\vert_{\ \verts{x}\ <\ 1} & = \sum_{n = 1}^{\infty}\overbrace{\braces{\sum_{i = 1}^{\infty}{\bracks{i \leq n} \over i}}}^{\ds{H_{n}}}\ \overbrace{\braces{\sum_{j = 1}^{\infty}{\bracks{j \leq n} \over j}}}^{\ds{H_{n}}}\ x^{n} \\[5mm] & = \sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j} \sum_{n = 1}^{\infty}\bracks{n \geq i}\bracks{n \geq j}x^{n} \\[5mm] & = \sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j}\braces{% \bracks{i \leq j}\sum_{n = j}^{\infty}x^{n} + \bracks{i > j}\sum_{n = i}^{\infty}x^{n}} \\[5mm] & = \sum_{i = 1}^{\infty}{1 \over i}\sum_{j = 1}^{\infty}{1 \over j}\braces{% \bracks{j \geq i}\,{x^{j} \over 1 - x} + \bracks{j < i}\,{x^{i} \over 1 - x}} \\[5mm] & = {1 \over 1 - x}\sum_{i = 1}^{\infty}{1 \over i}\pars{% \sum_{j = i}^{\infty}{x^{j} \over j} + \sum_{j = 1}^{i - 1}{x^{i} \over j}} = {1 \over 1 - x}\pars{\sum_{i = 1}^{\infty}{1 \over i} \sum_{j = i}^{\infty}{x^{j} \over j} + \sum_{i = 1}^{\infty}{x^{i} \over i}\overbrace{\sum_{j = 1}^{i - 1}{1 \over j}} ^{\ds{H_{i - 1}}}} \\[5mm] & = {1 \over 1 - x}\pars{\sum_{j = 1}^{\infty}{x^{j} \over j}\ \overbrace{\sum_{i = 1}^{j}{1 \over i}}^{\ds{H_{j}}}\ + \sum_{i = 1}^{\infty}{x^{i + 1} \over i + 1}\,H_{i}} = {1 \over 1 - x}\sum_{i = 1}^{\infty} \pars{{x^{i} \over i} + {x^{i + 1} \over i + 1}}H_{i} \\[5mm] & = {1 \over 1 - x}\sum_{i = 1}^{\infty}H_{i} \pars{x^{i}\int_{0}^{1}t^{i - 1}\,\dd t + x^{i + 1}\int_{0}^{1}t^{i}\,\dd t} \\[5mm] & = {1 \over 1 - x} \pars{\int_{0}^{1}{1 \over t}\sum_{i = 1}^{\infty}H_{i}\pars{xt}^{i}\,\dd t + x\int_{0}^{1}\sum_{i = 1}^{\infty}H_{i}\pars{xt}^{i}\,\dd t} \\[5mm] & = {1 \over 1 - x}\int_{0}^{1}{1 + xt \over t} \bracks{-\,{\ln\pars{1 - xt} \over 1 - xt}}\,\dd t = -\,{1 \over 1 - x}\int_{0}^{x}{1 + t \over t\pars{1 - t}}\, \ln\pars{1 - t}\,\dd t \end{align}

because $\ds{\sum_{i = 1}^{\infty}H_{i}z^{i} = -\,{\ln\pars{1 - z} \over 1 - z}:\ \pars{~H_{i}\ Generating\ Function~}}$.

Then, \begin{align} \left.\sum_{n = 1}^{\infty}H_{n}^{2}x^{n}\,\right\vert_{\ \verts{x}\ <\ 1} & = -\,{1 \over 1 - x}\bracks{% 2\ \underbrace{\int_{0}^{x}{\ln\pars{1 - t} \over 1 - t}\,\dd t} _{\ds{-\,{1 \over 2}\,\ln^{2}\pars{1 - x}}}\ +\ \underbrace{\int_{0}^{x}\overbrace{{\ln\pars{1 - t} \over t}} ^{\ds{-\,\mrm{Li}_{2}'\pars{x}}}\ \,\dd t} _{\ds{-\,\mrm{Li}_{2}\pars{x}}}} \\[5mm] & = \bbx{\ln^{2}\pars{1 - x} + \mrm{Li}_{2}\pars{x} \over 1 - x} \end{align}