TITLE: Stirling numbers and harmonic number identities
QUESTION [17 upvotes]: Permit me a brief introduction before I state the question, three questions in fact.
Inspired by this MSE link I computed the following harmonic sum identities:
$$1/6\, \left( {H_{{n}}}^{(1)}
\right) ^{3}-1/2\,{H_{{n}}}^{(2)}{H_{{n}}}^{(1)}+1/3\,{H_{{n}}}^{(3)}
= \frac{1}{n!} \left[n+1\atop 4\right]$$
and
$$1/24\,
\left( {H_{{n}}}^{(1)}
\right) ^{4}-1/4\,{H_{{n}}}^{(2)}
\left( {H_{{n}}}^{(1)}
\right) ^{2}+1/3\,{H_{{n}}}^{(3)}{H_{{n}}}^{(1)}
\\+1/8\,
\left( {H_{{n}}}^{(2)}
\right) ^{2}-1/4\,{H_{{n}}}^{(4)}
= \frac{1}{n!} \left[n+1\atop 5\right]$$
and finally
$$\frac{1}{120}{ \left( {H_{{n}}}^{(1)}
\right) ^{5}}
-1/12\,{H_{{n}}}^{(2)}
\left( {H_{{n}}}^{(1)}
\right) ^{3}+1/6\,{H_{{n}}}^{(3)}
\left( {H_{{n}}}^{(1)}
\right) ^{2}+1/8\,{H_{{n}}}^{(1)}
\left( {H_{{n}}}^{(2)}
\right) ^{2}\\-1/4\,{H_{{n}}}^{(4)}{H_{{n}}}^{(1)}
-1/6\,{H_{{n}}}^{(2)}{H_{{n}}}^{(3)}+1/5\,{H_{{n}}}^{(5)}
= \frac{1}{n!} \left[n+1\atop 6\right].$$
The parameter here call it $q$ takes on the values $q=3,4$ and $q=5.$

I verified the above identities numerically and they  seem to hold. The values of the
coefficients indicate that the  Polya Enumeration Theorem plays a role
here, and indeed  it is not difficult to  present a general conjecture
for the above pattern which is
$$\color{#A00}{\left.Z(P_q)
\left(\sum_{k=1}^n \frac{1}{k^{s}} \right)\right|_{s=1}
= \frac{1}{n!} \left[n+1\atop q+1 \right].}$$
We  use $Z(P_q)$  to refer  to the  cycle index  of the  unlabeled set
operator $\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}\textsc{SET}_{=q}$ (notation from Analytic Combinatorics by Flajolet and Sedgewick), which
admits the following simple recursive definition:
$$Z(P_0) = 1 \quad\text{and}\quad
Z(P_n) = \frac{1}{n}
\sum_{l=1}^n (-1)^{l+1} a_l Z(P_{n-l}).$$
Recall that we must distinguish between multisets ($\textsc{MSET}$) and sets ($\textsc{SET}$).
Here are the cycle indices $Z(P_3), Z(P_4)$ and $Z(P_5):$
$$\begin{array}{|l|l|}
\hline
Z(P_3) &
\frac{1}{6}\,{a_{{1}}}^{3}-1/2\,a_{{2}}a_{{1}}+1/3\,a_{{3}}\\
\hline
Z(P_4) &
\frac{1}{24}\,{a_{{1}}}^{4}-1/4\,a_{{2}}{a_{{1}}}^{2}
+1/3\,a_{{3}}a_{{1}}+1/8\,{a_{{2}}}^{2}-1/4\,a_{{
4}}\\
\hline
Z(P_5) &
{\frac {1}{120}}\,{a_{{1}}}^{5}-\frac{1}{12}\,a_{{2}}{a_{{1}}}^{3}
+1/6\,a_{{3}}{a_{{1}}}^{2}+1/8\,a_{{
1}}{a_{{2}}}^{2}-1/4\,a_{{4}}a_{{1}}
-1/6\,a_{{2}}a_{{3}}+1/5\,a_{{5}}\\
\hline\end{array}$$
The substitution mechanism for PET here is
$$a_l = \sum_{k=1}^n \frac{1}{k^{ls}}$$
and it should be immediate to see how the formulas were obtained
from the cycle indices.

The proof of  the conjecture now morphs into  a simple one-liner.  The
LHS is the ordinary generating function of sets having cardinality $q$
chosen  from  the  repertoire  $\{1, 1/2^s,  1/3^s,  \ldots,  1/n^s\}$
evaluated  at $s=1.$ By  inspection this  same generating  function is
also given by
$$[z^q] \prod_{k=1}^n \left(1 + \frac{z}{k^s}\right).$$
Putting $s=1$ we obtain
$$[z^q] \prod_{k=1}^n \left(1 + \frac{z}{k}\right)
= \frac{1}{n!} [z^q] \prod_{k=1}^n \left(z+k\right)
= \frac{1}{n!} [z^q] \frac{1}{z} \prod_{k=0}^n \left(z+k\right)
\\= \frac{1}{n!} [z^{q+1}] \prod_{k=0}^n \left(z+k\right)
= \frac{1}{n!} \left[ n+1\atop q+1 \right].$$

We have used the ordinary  generating function of the Stirling numbers
of the first kind in the last step of this calculation and obtained the RHS of
the conjectured formula which completes the proof.

These are the  three questions.  **First**,  I'd be  interested  in  possible
references  for this  formula, **second**, I'd  like  to know  if
perhaps it admits a more elementary proof, and **third** does this formula have computational uses e.g. it would appear to produce asymptotic expansions for certain Stirling numbers.

Thanks for reading.

Here is the Maple code that was used to verify the three formulas presented at the beginning. Even $q=25$ gives instant results. The combinatorics package is included for its Stirling number routines.

with(combinat);

pet_cycleind_set :=
proc(n)
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add((-1)^(l-1)*a[l]*pet_cycleind_set(n-l), l=1..n));
end;

v :=
proc(n, q)
    local sl, l;
    option remember;

    sl := [seq(a[l]=harmonic(n, l), l=1..q)];
    subs(sl, pet_cycleind_set(q));
end;

Addendum.
Here are the asymptotic expansions for the initial three identities:
$$\frac{1}{n!}\left[n+1\atop 4\right]
\sim 1/6\, \left( \ln  \left( n \right)  \right) ^{3}+1/2\, \left( \ln
 \left( n \right)  \right) ^{2}\gamma+ \left( 1/2\,{\gamma}^{2}-1/12
\,{\pi }^{2} \right) \ln  \left( n \right)\\ +1/6\,{\gamma}^{3}-1/12\,
{\pi }^{2}\gamma+1/3\,\zeta  \left( 3 \right)$$
and
$$\frac{1}{n!}\left[n+1\atop 5\right]
\sim 1/24\, \left( \ln  \left( n \right)  \right) ^{4}+1/6\,\gamma\,
 \left( \ln  \left( n \right)  \right) ^{3}\\+ \left( 1/4\,{\gamma}^{2
}-1/24\,{\pi }^{2} \right)  \left( \ln  \left( n \right)  \right) ^{
2}+ \left( 1/6\,{\gamma}^{3}-1/12\,{\pi }^{2}\gamma+1/3\,\zeta
 \left( 3 \right)  \right) \ln  \left( n \right) \\+1/24\,{\gamma}^{4}
-1/24\,{\pi }^{2}{\gamma}^{2}+1/3\,\zeta  \left( 3 \right) \gamma+{
\frac {{\pi }^{4}}{1440}}$$
and finally
$$\frac{1}{n!}\left[n+1\atop 6\right]
\sim {\frac { \left( \ln  \left( n \right)  \right) ^{5}}{120}}+1/24\,
\gamma\, \left( \ln  \left( n \right)  \right) ^{4}+ \left( 1/12\,{
\gamma}^{2}-{\frac {{\pi }^{2}}{72}} \right)  \left( \ln  \left( n
 \right)  \right) ^{3}\\+ \left( 1/12\,{\gamma}^{3}-1/24\,{\pi }^{2}
\gamma+1/6\,\zeta  \left( 3 \right)  \right)  \left( \ln  \left( n
 \right)  \right) ^{2}\\+ \left( 1/24\,{\gamma}^{4}-1/24\,{\pi }^{2}{
\gamma}^{2}+1/3\,\zeta  \left( 3 \right) \gamma+{\frac {{\pi }^{4}}{
1440}} \right) \ln  \left( n \right)\\ +{\frac {{\gamma}^{5}}{120}}-{
\frac {{\pi }^{2}{\gamma}^{3}}{72}}+1/6\,\zeta  \left( 3 \right) {
\gamma}^{2}+{\frac {\gamma\,{\pi }^{4}}{1440}}-1/36\,{\pi }^{2}
\zeta  \left( 3 \right) +1/5\,\zeta  \left( 5 \right).$$
The Maple code for these is as follows.

aex :=
proc(q)
    local sl, l;
    option remember;

    sl := [a[1]=log(n)+gamma,
           seq(a[l]=Zeta(l), l=2..q)];
    subs(sl, pet_cycleind_set(q));
end;

REPLY [2 votes]: By Taylor series, we have
$$\frac{\ln^q(1+x)}{1+x}=\sum_{n=0}^\infty \frac{q!}{n!}s(n+1,q+1)x^n$$
where $s(a,b)$ is the Stirling number of the first kind.
Replace $x$ by $-x$ then multiply both sides by $(-1)^q$ and use $s(a,b)=(-1)^{a-b}\left[a\atop b\right]$, where $\left[a\atop b\right]$ is the unsigned Stirling number of first kind, we have
$$(-1)^q\frac{\ln^q(1-x)}{1-x}=\sum_{n=1}^\infty \frac{q!}{n!}\left[n+1\atop q+1\right]x^n$$
On the other hand, we showed here
$$(-1)^q\frac{\ln^q(1-x)}{1-x}=\sum_{n=0}^\infty g(q,n) x^n,$$
$$g(q,n)=-(q-1)!\sum_{k=0}^{q-1}\frac{(-1)^{q-k}}{k!}g(k,n)H_n^{(q-k)},\quad g(0,n)=1.$$
Therefore
$$\frac{q!}{n!}\left[n+1\atop q+1\right]=g(q,n)=-(q-1)!\sum_{k=0}^{q-1}\frac{(-1)^{q-k}}{k!}g(k,n)H_n^{(q-k)},\quad g(0,n)=1.$$
Examples using Mathematica: