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 $f_n(n)$? 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...
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
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 \newcommand{\sech}{\,{\rm sech}}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

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}

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

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

A: 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:
$$ f_n(n) = [x^n]\left(\left(\sum_{k\geq 1}\frac{x^k}{k}\right)\cdot\left(\sum_{k\geq 0}\binom{n+k-1}{k}x^k\right)\right)=[x^n]\frac{-\log(1-x)}{(1-x)^n}. $$
