If $f_1(k)=\sum_{i=1}^k\frac{1}{i}$ and $f_n(k)=\sum_{i=1}^kf_{n-1}(i)$, then what is $f_n(n)$?

Question

Let $$f_1(k)=\sum_{i=1}^k\frac{1}{i},$$ and define inductively $$f_n(k)=\sum_{i=1}^kf_{n-1}(i).$$ So, $$f_2(k)=\sum_{i_2=1}^k\sum_{i_1=1}^{i_2}\frac{1}{i_1},\quad f_2(k)=\sum_{i_3=1}^k\sum_{i_2=1}^{i_3}\sum_{i_1=1}^{i_2}\frac{1}{i_1},$$ and so on.

Question: What is $f_n(n)$ for all $n\in\mathbb{N}$?

The first few terms are $$1,\frac{5}{2},\frac{47}{6},\frac{319}{12},\frac{1879}{20},\ldots$$ but I find difficult to find the general pattern.

Added: The numerators appear to be the coefficients in the power series of $$-\ln(1+x)\ln(1-x).$$ This is very interesting...

Answer 1

We have $$f_1(k)=H_k,\qquad f_2(k)=(k+1)H_k-k$$ and since: $$ f_n(n)=\sum_{k=1}^{n}f_{n-1}(k)=\sum_{k=1}^{n}\sum_{j=1}^{k}f_{n-2}(j)=\sum_{k=1}^{n}(n-k+1)\,f_{n-2}(k)=\sum_{k=1}^{n}\binom{n-k+2}{2}f_{n-3}(k)$$ we have: $$ f_n(n) = \sum_{k=1}^{n}\binom{n-k+n-1}{n-1}\frac{1}{k}=\sum_{k=1}^{n}\binom{2n-k-1}{n-1}\frac{1}{k}\\=\binom{2n-2}{n-1}\int_{0}^{1}\phantom{}_2 F_1(1,1-n;2-2n;x)\,dx$$ or: $$ f_n(n) = [x^n]\left(\left(\sum_{k\geq 1}\frac{x^k}{k}\right)\cdot\left(\sum_{k\geq 0}\binom{n+k-1}{k}x^k\right)\right)=[x^n]\frac{-\log(1-x)}{(1-x)^n}. $$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

With $\ds{\verts{z}\ <\ 1}$:

\begin{align} {\cal F}_{n}\pars{z}&\equiv\sum_{k\ =\ 1}^{\infty}\fermi_{n}\pars{k}z^{k} =\sum_{k\ =\ 1}^{\infty}\sum_{i\ =\ 1}^{k}\fermi_{n - 1}\pars{i}z^{k} =\sum_{i\ =\ 1}^{\infty}\fermi_{n - 1}\pars{i}\sum_{k\ =\ i}^{\infty}z^{k} =\sum_{i\ =\ 1}^{\infty}\fermi_{n - 1}\pars{i}\,{z^{i} \over 1 - z} \\[5mm]&={1 \over 1 - z}\sum_{i\ =\ 1}^{\infty}\fermi_{n - 1}\pars{i}z^{i} \end{align}

\begin{align} \imp&\quad{\cal F}_{n}\pars{z}={{\cal F}_{n - 1}\pars{z} \over 1 - z} ={{\cal F}_{n - 2}\pars{z} \over \pars{1 - z}^{2}}=\cdots ={{\cal F}_{1}\pars{z} \over \pars{1 - z}^{n - 1}} \\[5mm]&={1 \over \pars{1 - z}^{n - 1}}\, \sum_{k\ =\ 1}^{\infty}\ \overbrace{\fermi_{1}\pars{k}}^{\dsc{H_{k}}}z^{k} =-\,{1 \over \pars{1 - z}^{n - 1}}\,{\ln\pars{1 - z} \over 1 - z} =-\,{\ln\pars{1 - z} \over \pars{1 - z}^{n}} \\[5mm]&=-\lim_{\mu\ \to\ -n}\partiald{\pars{1 - z}^{\mu}}{\mu} =-\lim_{\mu\ \to\ -n}\partiald{}{\mu} \sum_{k\ =\ 0}^{\infty}{\mu \choose k}\pars{-1}^{k}z^{k} \\[5mm]&=-\lim_{\mu\ \to\ -n}\partiald{}{\mu} \sum_{k\ =\ 1}^{\infty}{-\mu + k - 1\choose k}z^{k}\quad\imp\quad \fermi_{n}\pars{k}=-\lim_{\mu\ \to\ -n}\partiald{}{\mu} {-\mu + k - 1\choose k} \end{align}

$$ \fermi_{n}\pars{k}={\Gamma\pars{k + n} \over \Gamma\pars{k + 1}\Gamma\pars{n}}\, \bracks{\Psi\pars{k + n} - \Psi\pars{n}} $$

$$ \color{#66f}{\large\fermi_{n}\pars{n}} ={\Gamma\pars{2n} \over \Gamma\pars{n + 1}\Gamma\pars{n}}\, \bracks{\Psi\pars{2n} - \Psi\pars{n}} =\color{#66f}{\large{2n - 1 \choose n}\bracks{\Psi\pars{2n} - \Psi\pars{n}}} $$