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This is a natural extension of the question Sum of Squares of Harmonic Numbers. I became interested in this question while studying the problem A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$, especially the variation with a general integer power $q$, of which the present question is a part.

Let the sum in question be

$$s(q,n) = \sum _{k=1}^n H_k^{q} \tag{1}$$

where $q = 1, 2, 3, ...$ and $H_k = 1 + 1/2 + ... + 1/k$ is the harmonic number.

Problems

  1. Derive a recursion relation for $s(q,n)$
  2. Calculate $s(q,n)$ for $q=1 .. 4$
  3. Discuss similarities with and possible deviation from the corresponding integrals $i(q,n)=\int_1^n ln(k)^q \, dk$
  4. Discuss the possible meaning of the term "closed form" in the present context of finite sums.

I have given an answer to 1. and 2. up to q = 4, and 3. correspondingly.

Question 4. is perhaps the most interesting, and I have started to answer it.

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The sum for which we wish to find a closed expression is

$$s(q,n) = \sum _{k=1}^n H_k^{q} \tag{1}$$

where $q = 1, 2, 3, ...$ and $H_k = 1 + 1/2 + ... + 1/k$ is the harmonic number.

1. Recursion relation

$s(q,n)$ satisfies this recursion relation

$$s(q,n) = H_n s(q-1,n) - \sum _{m=1}^n \frac{1}{m} s(q-1,m-1) \tag{2}$$

with the boundary conditions

$$s(0,n)=n, s(q,0)=0 \tag{2a}$$

Derivation

Splitting one factor $H_k$ off from the power we can write

$$s(q,n) = \sum _{k=1}^n H_k^{q} = \sum _{k=1}^n H_k^{q-1}\;H_k$$

Inserting $H_k = \sum _{m=1}^k \frac{1}{m}$ this becomes

$$s(q,n) = \sum _{k=1}^n H_k^{q-1}\sum _{m=1}^k \frac{1}{m}$$

Exchanging the order of summation gives

$$s(q,n) =\sum _{m=1}^n \frac{1}{m} \sum _{k=m}^n H_k^{q-1}$$

This can be written as

$$s(q,n) =\sum _{m=1}^n \frac{1}{m} (\sum _{k=1}^n H_k^{q-1} - \sum _{k=1}^{m-1} H_k^{q-1})$$

which using $(1)$ gives $(2)$.

Notice that this derivation is different from Abelian partial summation.

The boundary conditions follow directly from the definition $(1)$.

2. The sum for particular values of q

We also give the corresponding integral over the power of ln(n) (see 3. below).

From $(2)$ we obtain easily

$$s(1,n)=(n+1) H_n-n \tag{3}$$ $$i(1,n)=n \ln (n) -n +1 \tag{3a}$$ and

$$s(2,n)=(n+1) H_n^{2}-(2 n+1) H_n+2 n \tag{4}$$ $$i(2,n)=n \ln ^2(n)-2 n \ln (n) +2 n-2 \tag{4a}$$

Slightly more effort is required for $q = 3$ which gives

$$s(3,n) = (n+1) H_n^{3}-(3 n+1) H_n^{2}+(6 n+3) H_n-6 n - \frac{1}{2} \left( H_n^{2}-H_n^{(2)} \right)\tag{5}$$ $$i(3,n)=n \ln ^3(n)-3 n \ln ^2(n)+6 n \ln (n) - 6 n+6 \tag{5a}$$

where we encounter the harmonic sum of second order

$$H_n^{(2)} = \sum _{m=1}^n \frac{1}{m^{2}} \tag{5b}$$

and the curious term

$$\sum _{k=1}^n \frac{H_{k-1}}{k}=\frac{1}{2} \left(H_n^2-H_n^{(2)}\right)\tag{5c}$$

which is also the sum of the terms $1/(i\; j)$ above the diagonal, i.e. for $1\le i<j\le n$.

In the case $q=4$ we find (notice that (6) is preliminary, see (6e) below for a more satisfactory result)

$$s(4,n) =(n+1) H_n^4-\left(4 n+\frac{3}{2}\right) H_n^3+(12 n+6) H_n^2-(24 n+12) H_n+24 n+\left(\frac{1}{2}H_n-2\right) H_n^{(2)}-\frac{3}{2}s_{41}(n)-\frac{1}{2}s_{42}(n)\tag{6}$$

$$i(4,n) = n \ln ^4(n)-4 n \ln ^3(n)+12 n \ln ^2(n)-24 n \ln (n) + 24 n-24\tag{6a}$$

We encouter two new sums which we express via some "canonical" forms $h_i$of sums

$$s_{41}(n)=\sum _{m=1}^n \frac{H_{m-1}^2}{m} = h_1(n) - 2 h_2(n) + h_3(n)\tag{6c}$$

$$s_{42}(n)=\sum _{m=1}^n \frac{H_{m-1}^{(2)}}{m} = h_4(n) - h_3(n)\tag{6d}$$

Where

$$h_1(n) = \sum _{k=1}^n \frac{H_k^2}{k}\tag{7a}$$ $$h_2(n) = \sum _{k=1}^n \frac{H_k}{k^2}\tag{7b}$$ $$h_3(n) = \sum _{k=1}^n \frac{1}{k^3}\tag{7c}$$ $$h_4(n)=\sum _{k=1}^n \frac{H_k^{(2)}}{k}\tag{7d}$$

We have

$$h_3(n) = H_n^{(3)}\tag{7e}$$

and it was shown in [1] that

$$h_1(n) - h_2(n) = \frac{1}{3} (H_n^3 - H_n^{(3)})\tag{8a}$$

and

$$h_4(n)=-h_2(n)+H_n H_n^{(2)}+H_n^{(3)}\tag{8b}$$

so that only one of the three sums $h_1$, $h_2$, and $h_4$ is independent.

Selecting $h_2$ as the independent new term we obtain for $s(4,n)$ the simplified expression

$$s(4,n)=(n+1) H_n^4-(4 n+2) H_n^3+(12 n+6) H_n^2-(24 n+12) H_n+24 n\;+(2 h_2(n)-2 H_n^{(2)}-H_n^{(3)})\tag{6e}$$

Notice that the additional term in brackets is bounded for $n\to\infty$.
In fact, it is easy to see (using robjohn's result for $h_2$ in the solution to [1]) that

$$\lim_{n\to \infty } \, (2 h_2(n)-2 H_n^{(2)}-H_n^{(3)})=4 \zeta(3) -2\zeta(2)-\zeta(3) = 3\zeta(3)-2\zeta(2)=0.316303...\tag{6f}$$

Derivation

The derivation of the formulae for the sum of the powers is straightforward using the recursion relation but it is tedious and must be done carefully.

(8a) was derived in [1]

(8b) is easy to prove by induction. The derivation, however, is more laborious and will be given here using the method of exchanging the order of summation in the double sum which is obtained after entering the definition of $H_k^{(2)}$

$$ \begin{align} \sum _{k=1}^n \frac{H_k^{(2)}}{k} &=\frac{1}{1} (1)\tag{9a}\\ &+ \frac{1}{2}(1+\frac{1}{2^2})\\ &+ \frac{1}{3}(1+\frac{1}{2^2}+\frac{1}{3^2})\\ &+ ... \\ &+ \frac{1}{n}(1+\frac{1}{2^2}+\frac{1}{3^2}+ ...+\frac{1}{n^2} )\\ &=\frac{1}{1^2} H_n +\frac{1}{2^2}(H_n-H_1)+\frac{1}{3^2}(H_n-H_2)+ ... +\frac{1}{n^2}(H_n-H_{n-1}) \tag{9b}\\ &=\sum _{k=1}^{n} \frac{1}{k^2}(H_n-H_{k-1})\\ &=H_n^{(2)}H_n - \sum _{k=1}^n \frac{1}{k^2} (H_k-\frac{1}{k})= H_n^{(2)}H_n-h_2(n) + H_n^{(3)}\tag{9c}\\[9pt] \end{align} $$

Explanation:
$\text{(9a)}$: replace $H_k^{(2)}$ by it definition (5b) an write down the sum like in the following lines
$\text{(9b)}$: Exchange the order of summation, i.e. sum vertically, and take care of the triangular shape of the array of summands
$\text{(9c)}$: identify known sums, let $H_{k-1} = H_k -\frac{1}{k}$, expand, and identify knows sums again.
Derivation of $\text{(8b)}$ completed.

3. The integral of powers of $ln(n)$

Jack d'Aurizio observed in [2] that the integrals

$$i(q,n)=\int_1^n ln(k)^q \, dk\tag{10}$$

(for $q = 1$ and $q = 2$) have a close similarity to the corresponding sums.

In fact

$$i(1,n)=n \ln (n) -n +1 \tag{3a}$$ $$i(2,n)=n \ln ^2(n)-2 n \ln (n) +2 n-2 \tag{4a}$$ $$i(3,n)=n \ln ^3(n)-3 n \ln ^2(n)+6 n \ln (n) - 6 n+6 \tag{5a}$$ $$i(4,n) = n \ln ^4(n)-4 n \ln ^3(n)+12 n \ln ^2(n)-24 n \ln (n) + 24 n-24\tag{6a}$$

It is not difficult to see that in general

$$i(k,n)=(-1)^{m+1} m!+\sum _{k=0}^m (-1)^k n \binom{m}{k} \ln^{m-k}(n)\tag{11}$$

It is interesting that already for $q=3$ this similarity is "spoiled" by the term (5c)

For $q=4$ other types of "spoiler" terms appear (see above).

4. The meaning of "closed form"

As we are dealing with finite sums here these could themselves be called closed forms. But, of course we want to have something "shorter" than the original expression.

We can already see some structure in the cases up to the fourth power: guided by the correponding integral we find for $s(q,n)$ a linear combination of all powers of $H_n$ from zero to $q+1$ the coefficients are linear in $n$. Starting from $q=3$, however, "new" types of sums appear: harmonic numbers of higher order and additional "correction" terms (relative to the ln-integral)which are still under study here.

We could tentatively define "closed expression" here as being a "short" linear combination of expressions within a (hopefully) finite class $v$ of sums, we can call basic elements.

Specific question: Is it sufficient to have these two basic elements

$$v = (H_{n}, H_{n}^{(m)})$$

or is, for instance $h_1$ or, equivalently $h_2$ or $h_4$

also a basic element which must be added to $v$?

5. References

$\text[1]$ Is there a closed form for $\sum _{k=1}^n \frac{1}{k}H_{k-1}^2$?

$\text[2]$ A closed form of $\sum_{n=1}^\infty\left[ H_n^2-\left(\ln n+\gamma+\frac1{2n} \right)^2\right]$

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I'm coming back to this question after almost one year because I was able to find the complete answer.

Recently, I became aware of the existence of the summation tool Sigma.m which is available as a Mathematica package (https://risc.jku.at/sw/sigma/).

This tool provides a framework for relating (harmonic) sums of different types, and I used it heuristically for the present task. Of course, also with this tool there is no "free dinner". I had to study several special cases, find a pattern and then make a good guess for the set of basic sums through wich a given sum can be expressed. The guess was then confirmed for a number of cases.

Main result

The complete answer to the question can be formulated as a follows

The finite sum of the q-th power of the harmonic number

$$s_{q}(n) = \sum_{k=1}^n H_{k}^q,\;\; q = 1, 2, 3...\tag{1}$$

can be epressed in terms of a set of basic sums

$$t_{q} = \{H_n,\{H_{n}^{(r)},\; r=1..q\}, \{\sum_{k=1}^n \frac{H_{k}^{(r)}}{k}, \;{r=2..q}\},\\\{\sum_{k=1}^n \frac{H_{k}^r}{k^s},\;r+s=3..q-1,\; r,s,\ge 1,\; q\ge 4\}\}\tag{2}$$

Where, for given $q$, all basic sums whose parameters obey the ranges indicated must be taken into account.

This set is called "Tower" in the framework of Sigma. The main task was to find the correct form of the tower.

Proof

Up to now I have no formal proof. However, as the particular results for q=4 through q=10 are correct, I see it as a strong indication that the theorem is valid. The next task would be to find the expressions for the coefficients.

Technical remarks

For Mathematica users here's the code for the Tower

tow[q_] := Join[{SigmaHNumber[n]},
  Table[SigmaHNumber[m, n], {m, 2, q}],
  Table[SigmaSum[1/k SigmaHNumber[m, k], {k, 1, n}], {m, 2, q - 2}],
  Table[SigmaSum[1/k^m SigmaHNumber[k]^(q - 1 - m), {k, 1, n}], {m, 1,
     q - 2}],
  Flatten[
   Table[Table[
     SigmaSum[1/k^m SigmaHNumber[k]^(p + 1 - m), {k, 1, n}], {m, 1, 
      p}], {p, 2, q - 2}]]]

The outpunt $f$ of Sigma was transformed to Mathematica (and then to LaTEX) using the replacements

rep[f_] := 
 f /. Subscript[\[Iota], 1] -> k /. 
   Sigma`Summation`Objects`Private`MySum -> Sum /. 
  Sigma`Summation`Objects`Private`HNumber[a_, b_] -> 
   HarmonicNumber[b, a]

and xSum was subsequently replaced textually with Sum, finally LaTEX was taken from that expression. The use of xSum was necessary to prevent Mathematica from (prematurely) calculating the expressions.

The sums for $q=3..10$

Here comes the output of Sigma with the tower (2). Sorry for not having it "beautified" with respect to the order of the terms for now.

Notice that the cases $q=3$ und $q=4$ confirm the results obatained earlier "manually".

$$s_{3} = (n+1) \left(H_n\right){}^3-\frac{3}{2} (2 n+1) \left(H_n\right){}^2+3 (2 n+1) H_n+\frac{H_n^{(2)}}{2}-6 n$$

$$s_{4}=-2 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+H_n \left(2 H_n^{(2)}-12 (2 n+1)\right)+(n+1) \left(H_n\right){}^4-2 (2 n+1) \left(H_n\right){}^3+6 (2 n+1) \left(H_n\right){}^2-2 H_n^{(2)}+H_n^{(3)}+24 n$$

$$s_{5}=-\frac{10}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+10 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+\frac{5}{3} \sum _{k=1}^n \frac{H_k^{(3)}}{k}+H_n \left(-10 H_n^{(2)}-\frac{5 H_n^{(3)}}{3}+60 (2 n+1)\right)+(n+1) \left(H_n\right){}^5-\frac{5}{3} (3 n+2) \left(H_n\right){}^4+10 (2 n+1) \left(H_n\right){}^3-30 (2 n+1) \left(H_n\right){}^2+10 H_n^{(2)}-5 H_n^{(3)}-H_n^{(4)}-120 n$$

$$s_{6}=-5 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-20 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+\frac{15}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-60 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-10 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-\frac{3}{2} \sum _{k=1}^n \frac{H_k^{(4)}}{k}+H_n \left(60 H_n^{(2)}+10 H_n^{(3)}+\frac{3 H_n^{(4)}}{2}-360 (2 n+1)\right)+(n+1) \left(H_n\right){}^6-\frac{3}{2} (4 n+3) \left(H_n\right){}^5+10 (3 n+2) \left(H_n\right){}^4-60 (2 n+1) \left(H_n\right){}^3+180 (2 n+1) \left(H_n\right){}^2-60 H_n^{(2)}+30 H_n^{(3)}+6 H_n^{(4)}+H_n^{(5)}+720 n$$

$$s_{7} = 7 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+35 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-14 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}+140 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}-\frac{105}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}+\frac{63}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+420 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+70 \sum _{k=1}^n \frac{H_k^{(3)}}{k}+\frac{21}{2} \sum _{k=1}^n \frac{H_k^{(4)}}{k}+\frac{7}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}+H_n \left(-420 H_n^{(2)}-70 H_n^{(3)}-\frac{21 H_n^{(4)}}{2}-\frac{7 H_n^{(5)}}{5}+2520 (2 n+1)\right)+(n+1) \left(H_n\right){}^7-\frac{7}{5} (5 n+4) \left(H_n\right){}^6+\frac{21}{2} (4 n+3) \left(H_n\right){}^5-70 (3 n+2) \left(H_n\right){}^4+420 (2 n+1) \left(H_n\right){}^3-1260 (2 n+1) \left(H_n\right){}^2+420 H_n^{(2)}-210 H_n^{(3)}-42 H_n^{(4)}-7 H_n^{(5)}-H_n^{(6)}-5040 n$$

$$s_{8} = -\frac{28}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}-56 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+\frac{70}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}-280 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}+112 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}-28 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}-1120 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+420 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-\frac{504}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+\frac{56}{3} \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}-3360 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-560 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-84 \sum _{k=1}^n \frac{H_k^{(4)}}{k}-\frac{56}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}-\frac{4}{3} \sum _{k=1}^n \frac{H_k^{(6)}}{k}+H_n \left(3360 H_n^{(2)}+560 H_n^{(3)}+84 H_n^{(4)}+\frac{56 H_n^{(5)}}{5}+\frac{4 H_n^{(6)}}{3}-20160 (2 n+1)\right)+(n+1) \left(H_n\right){}^8-\frac{4}{3} (6 n+5) \left(H_n\right){}^7+\frac{56}{5} (5 n+4) \left(H_n\right){}^6-84 (4 n+3) \left(H_n\right){}^5+560 (3 n+2) \left(H_n\right){}^4-3360 (2 n+1) \left(H_n\right){}^3+10080 (2 n+1) \left(H_n\right){}^2-3360 H_n^{(2)}+1680 H_n^{(3)}+336 H_n^{(4)}+56 H_n^{(5)}+8 H_n^{(6)}+H_n^{(7)}+40320 n$$

$$s_{9} = 12 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^5}+84 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}-36 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^4}+504 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}-210 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}+54 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^3}+2520 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}-1008 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}+252 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}-48 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^2}+10080 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}-3780 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}+\frac{4536}{5} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}-168 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}+\frac{180}{7} \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k}+30240 \sum _{k=1}^n \frac{H_k^{(2)}}{k}+5040 \sum _{k=1}^n \frac{H_k^{(3)}}{k}+756 \sum _{k=1}^n \frac{H_k^{(4)}}{k}+\frac{504}{5} \sum _{k=1}^n \frac{H_k^{(5)}}{k}+12 \sum _{k=1}^n \frac{H_k^{(6)}}{k}+\frac{9}{7} \sum _{k=1}^n \frac{H_k^{(7)}}{k}+H_n \left(-30240 H_n^{(2)}-5040 H_n^{(3)}-756 H_n^{(4)}-\frac{504 H_n^{(5)}}{5}-12 H_n^{(6)}-\frac{9 H_n^{(7)}}{7}+181440 (2 n+1)\right)+(n+1) \left(H_n\right){}^9-\frac{9}{7} (7 n+6) \left(H_n\right){}^8+12 (6 n+5) \left(H_n\right){}^7-\frac{504}{5} (5 n+4) \left(H_n\right){}^6+756 (4 n+3) \left(H_n\right){}^5-5040 (3 n+2) \left(H_n\right){}^4+30240 (2 n+1) \left(H_n\right){}^3-90720 (2 n+1) \left(H_n\right){}^2+30240 H_n^{(2)}-15120 H_n^{(3)}-3024 H_n^{(4)}-504 H_n^{(5)}-72 H_n^{(6)}-9 H_n^{(7)}-H_n^{(8)}-362880 n$$

$$s_{10} =-15 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^6}-120 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^5}+\frac{105}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^5}-840 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^4}+360 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^4}-\frac{189}{2} \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^4}-5040 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^3}+2100 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^3}-540 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^3}+105 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^3}-25200 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k^2}+10080 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k^2}-2520 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k^2}+480 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k^2}-75 \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k^2}-100800 \sum _{k=1}^n \frac{\left(H_k\right){}^3}{k}+37800 \sum _{k=1}^n \frac{\left(H_k\right){}^4}{k}-9072 \sum _{k=1}^n \frac{\left(H_k\right){}^5}{k}+1680 \sum _{k=1}^n \frac{\left(H_k\right){}^6}{k}-\frac{1800}{7} \sum _{k=1}^n \frac{\left(H_k\right){}^7}{k}+\frac{135}{4} \sum _{k=1}^n \frac{\left(H_k\right){}^8}{k}-302400 \sum _{k=1}^n \frac{H_k^{(2)}}{k}-50400 \sum _{k=1}^n \frac{H_k^{(3)}}{k}-7560 \sum _{k=1}^n \frac{H_k^{(4)}}{k}-1008 \sum _{k=1}^n \frac{H_k^{(5)}}{k}-120 \sum _{k=1}^n \frac{H_k^{(6)}}{k}-\frac{90}{7} \sum _{k=1}^n \frac{H_k^{(7)}}{k}-\frac{5}{4} \sum _{k=1}^n \frac{H_k^{(8)}}{k}+H_n \left(302400 H_n^{(2)}+50400 H_n^{(3)}+7560 H_n^{(4)}+1008 H_n^{(5)}+120 H_n^{(6)}+\frac{90 H_n^{(7)}}{7}+\frac{5 H_n^{(8)}}{4}-1814400 (2 n+1)\right)+(n+1) \left(H_n\right){}^{10}-\frac{5}{4} (8 n+7) \left(H_n\right){}^9+\frac{90}{7} (7 n+6) \left(H_n\right){}^8-120 (6 n+5) \left(H_n\right){}^7+1008 (5 n+4) \left(H_n\right){}^6-7560 (4 n+3) \left(H_n\right){}^5+50400 (3 n+2) \left(H_n\right){}^4-302400 (2 n+1) \left(H_n\right){}^3+907200 (2 n+1) \left(H_n\right){}^2-302400 H_n^{(2)}+151200 H_n^{(3)}+30240 H_n^{(4)}+5040 H_n^{(5)}+720 H_n^{(6)}+90 H_n^{(7)}+10 H_n^{(8)}+H_n^{(9)}+3628800 n $$

Discussion

First of all, the strict proof would be desirable. I invite others herewith to contribute.

There are several natural questions emerge from this result are, for instance:

a) is the set $t_{q}$ minimal?

b) are the sums in $t_{q}$ really basic or can the be further reduced?

c) are there equiavalent sets to (2)? We have already confirmed this for $q=4$ in the first answer.

d) can we find explicit expressions for the coefficents and hence construct the explicit closed expression for $s_{q}$? Note that the number of basic sums in $s_{q}$ is given by $\frac{1}{2} (q-2)(q-3)$.

Generalizations could include modified harmonic sums.

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