Double harmonic sum $\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$

Question

Are there any general formula for the following series

$$\tag{1}\sum_{n\geq 1}\frac{H^{(p)}_nH_n}{n^q}$$

Where we define

$$H^{(p)}_n= \sum_{k=1}^n \frac{1}{k^p}\,\,\,\,\,H^{(1)}_n\equiv H_n =\sum_{k=1}^n\frac{1}{k} $$

For the special case $p=q=2$ in (1) I found the following paper

Stating that

$$\sum_{n\geq 1}\frac{H^{(2)}_nH_n}{n^2}=\zeta(5)+\zeta(2)\zeta(3)$$

See equation (3a) .

Is there any other paper in the literature discussing (1) or any special cases ?

Answer 1

A couple of places to start for further looking:

The paper Further summation formulae related to generalized harmonic numbers by Zheng mentions the $p = q = 2$ case in example 2.3, and has several related results, but not explicitly of the form you're looking for.

Appendix B of the paper On some log-cosine integrals related to $\zeta(3), \zeta(4)$ and $\zeta(6)$ by Mark Coffey has some similar things, and maybe looking through its references and papers that cite it would yield more.

Answer 2

Here we provide an answer for $p=1$ and arbitrary $q \ge 2$. Let us denote the quantity in question as follows: \begin{equation} {\mathfrak S}^{(p,1)}_q := \sum\limits_{m=1}^\infty \frac{H^{(p)}_m H_m}{m^q} \end{equation} In order to calculate the above we start from the generating function. We have: \begin{eqnarray} \sum\limits_{m=1}^\infty H^{(p)}_m H_m \frac{t^m}{m} &= & - \int\limits_0^t \frac{\log(1-\xi/t)}{\xi} \cdot \frac{Li_p(\xi)}{1-\xi} d\xi \\ &=& \left\{ \begin{array}{rr} -\frac{1}{3} [\log(1-t)]^3 -\log(1-t) Li_2(t) + Li_3(t) & \mbox{for $p=1$} \\ \cdots \end{array} \right. \end{eqnarray} Now we use the common trick,i.e. we divide by $t$ and multiply by an appropriate power of $\log(1/t)$ and integrate appropriately. We have: \begin{eqnarray} {\mathfrak S}^{(1,1)}_q &=& \underbrace{-\frac{1}{3} \int\limits_0^1 \frac{[\log(1/\xi)]^{q-2}}{(q-2)!} \cdot \frac{[\log(1-\xi)]^3}{\xi} d\xi}_{I_1} +\\ &&\underbrace{\int\limits_0^1 \frac{[\log(1/\xi)]^{q-2}}{(q-2)!} \cdot \frac{Li_1(\xi) Li_2(\xi)}{\xi} d\xi}_{I_2}+\\ &&\underbrace{\int\limits_0^1 \frac{[\log(1/\xi)]^{q-2}}{(q-2)!} \cdot \frac{ Li_3(\xi)}{\xi} d\xi}_{I_3} \end{eqnarray} Clearly all the inthree integrals in here have already been dealt with in MSE. The last integral is trivial . We have: \begin{equation} I_3= Li_{q+2}(+1) \end{equation} The second integral has been evaluated in An integral involving product of poly-logarithms and a power of a logarithm.. We have: \begin{equation} I_2= \frac{1}{2}\left[ \zeta(2) \zeta(q) + \sum\limits_{j=1}^{q-3} j \zeta(q-j-1) \zeta(j+3)-\sum\limits_{j=1}^{q-2} j {\bf H}^{(q-j-1)}_{j+3}(+1) \right] \end{equation} where the quantities highlighted in bold font have been calculated in Calculating alternating Euler sums of odd powers .They all reduce to single zeta values for $q+2\le 7$ and otherwise involve a few additional two-dimensional zeta values. Finally the first integral has been evaluated in Compute an integral containing a product of powers of logarithms. .It always reduces to single zeta values. We have: \begin{eqnarray} &&I_1 = \\ &&\frac{(-1)^{q-1}}{(q-1)!} \left[ -\frac{1}{3} \Psi^{(q+1)}(1) + \frac{1}{2} \sum\limits_{j=1}^{q-2} \binom{q-1}{j} \left\{\Psi^{(j+1)}(1)\Psi^{(q-1-j)}(1) + \Psi^{(j+0)}(1)\Psi^{(q-j)}(1)\right\}+\right.\\ && \left. -\frac{1}{3} \sum\limits_{1 \le j < j_1 \le q-2} \frac{(q-1)!}{j! (j_1-j)!(q-1-j_1)!} \Psi^{(j)}(1) \Psi^{(j_1-j)}(1) \Psi^{(q-1-j_1)}(1) \right] \end{eqnarray} where $\Psi^{(j)}(1)= (-1)^{j+1} j! \zeta(j+1)$ for $j=1,2,3,\cdots$.

Answer 3

In here we provide a generating function of the quantities in question. Let us define: \begin{equation} {\bf H}^{(p,r)}_q(t) := \sum\limits_{m=1}^\infty H_m^{(p)} H_m^{(r)} \frac{t^m}{m^q} \end{equation} In here we take $q\ge1$. We have: \begin{eqnarray} &&{\bf H}^{(p,1)}_q(t) = Li_p(1) \cdot \frac{1}{2} [\log(1-t)]^2 \cdot 1_{q=1}+\\ &&\frac{(-1)^{q}}{2} \sum\limits_{l=(q-2)}^{p+q-3} \left(\binom{l}{q-2} 1_{l < p+q-3} + ({\mathcal A}^{(p)}_{q-2}) 1_{l=p+q-3}\right) \cdot \underbrace{\int\limits_0^1 \frac{[Li_1(t \xi)]^2}{\xi} Li_{l+1}(\xi) \frac{[\log(1/\xi)]^{p+q-3-l}}{(p+q-l-3)!}d\xi}_{I_1}+\\ && \frac{(-1)^{q-1}}{2} \sum\limits_{j=0}^{q-3} \left({\mathcal A}^{(p)}_{q-2-j}\right) \cdot \zeta(p+q-2-j) \underbrace{\int\limits_0^1 \frac{[Li_1(t \xi)]^2}{\xi} \frac{[\log(\xi)]^j}{j!} d\xi}_{I_2}+\\ && \sum\limits_{l=1}^p \underbrace{\int\limits_0^1 \frac{Li_q(t \xi)}{\xi} Li_l(\xi) \frac{[\log(1/\xi)]^{p-l}}{(p-l)!} d\xi}_{I_3} \end{eqnarray} Here $t\in (-1,1)$ and $p=1,2,\cdots$ and \begin{equation} {\mathcal A}^{(p)}_{q} := p+\sum\limits_{j=2}^{q} \binom{p+j-2}{j} = p \cdot 1_{p=1} + \frac{p+q-1}{p-1} \binom{p+q-2}{q}\cdot 1_{p > 1} \end{equation}

Note 1: The quantities in the right hand side all contain products of poly-logarithms and a power of logarithm. Those quantities, in principle, have been already dealt with in An integral involving product of poly-logarithms and a power of a logarithm. for example.

Note 2: Now that we have the generating functions we will find recurrence relations for the sums in question and hopefully provide some closed form expressions .

Now we have: \begin{eqnarray} &&I_1 =\\ &&\sum\limits_{l_1=2}^{l+1} \binom{p+q-2-l_1}{l+1-l_1} (-1)^{l+1-l_1} \zeta(l_1) \left({\bf H}^{(1)}_{p+q-l_1}(t) - Li_{p+q+1-l_1}(t) \right)+\\ &&\sum\limits_{l_1=2}^{p+q-2-l} \binom{p+q-2-l_1}{l} (-1)^{l-1} \zeta(l_1) \left({\bf H}^{(1)}_{p+q-l_1}(t) - Li_{p+q+1-l_1}(t) \right)+\\ &&\sum\limits_{l_1=1}^{p+q-2-l} \binom{p+q-2-l_1}{l} (-1)^{l-0} \left( {\bf H}^{(l_1,1)}_{p+q-l_1}(t) - {\bf H}^{(l_1)}_{p+q+1-l_1}(t) \right) \end{eqnarray} and \begin{eqnarray} &&I_2=2 (-1)^j \left[ {\bf H}^{(1)}_{j+2}(t) - Li_{j+3}(t)\right] \end{eqnarray} and \begin{eqnarray} &&I_3=\\ &&\sum\limits_{l_1=2}^l \binom{p-l_1}{p-l}(-1)^{l-l_1} \zeta(l_1) Li_{p+q+1-l_1}(t) +\\ && \sum\limits_{l_1=2}^{p-l+1} \binom{p-l_1}{l-1}(-1)^{l} \zeta(l_1) Li_{p+q+1-l_1}(t)- \sum\limits_{l_1=1}^{p-l+1} \binom{p-l_1}{l-1} (-1)^l {\bf H}^{(l_1)}_{p+q+1-l_1}(t) \end{eqnarray}