Finite sums of products of harmonic numbers like $\sum_{k=1}^n H_{k} H_{2k}$

Question

The problem

In Sum of powers of Harmonic Numbers finite sums of powers of the same harmonic number have been studied:

$$s(q,n) = \sum_{k=1}^n H_{k}^q\tag{1}$$

Here we turn to the related questions of calculating finite sums of products of harmonic numbers of the type

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

One interesting case is $p=1, q=2$

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

Another one is

$$s_{2}(1,4,n)=\sum_{k=1}^n H_{ k} H_{4 k} \tag{4}$$

We also ask for

$$s_{2}(0,2,n)=\sum_{k=1}^n H_{2 k} \tag{5}$$

$$s_{2}(2,2,n)=\sum_{k=1}^n H_{2 k}^2 \tag{6}$$

I found already (3) a challenge. I tried the usual techniques like partial summation and interchanging the order of summation but did not yet find a satisfactory result (see my self answer).

Questions

1) Can you calculate (3) through (6)?

2) Did you find techniques appropriate to more general cases of $p,q$?

@ joriki Thanks for the hint, corrected. I took the opportunity to extend the question to even simpler cases. — Dr. Wolfgang Hintze, Sep 08 '18 at 14:34
What is the issue in managing $\sum_{k=1}^{n}H_k H_{2k}$ by summation by parts? — Jack D'Aurizio, Sep 08 '18 at 15:51
@ Jack D'Aurizio you can see the issue now in my self answer. — Dr. Wolfgang Hintze, Sep 08 '18 at 20:45
There is software for these kinds of questions by Carsten Schneider: https://risc.jku.at/m/carsten-schneider — Mark, Sep 09 '18 at 19:55

Dr. Wolfgang Hintze · Accepted Answer · 2018-09-08T20:40:12.010

This is to track the progress.

As to (5): done

Mathematica gives

$\sum _{k=1}^n H_{2 k} = (n+1) H_{2 (n+1)}- n-\frac{3}{2} +\frac{1}{4}\left(-\psi ^{(0)}\left(n+\frac{3}{2}\right)+\psi ^{(0)}\left(\frac{3}{2}\right)\right)\tag{5a}$

or

$\sum _{k=1}^n H_{2 k} = (n+1) H_{2 (n+1)}- n-\frac{3}{2} -\frac{1}{4}\left( H_{n+\frac{1}{2}}-H_{\frac{1}{2}}\right)\tag{5b}$

As to (3): one sum open

Partial summation differentiating with respect to $H_{2k}$ led me to

$$\sum _{k=1}^n H_k H_{2 k} = A(n) H_{2 n}-A(n-1)+p_{1}(n)-\frac{1}{2} p_{2}(n-1)\tag{3a}$$

where

$A(n)=\sum_{k=1}^n H_{k} = (n+1)H_{n}-n\tag{3b}$

$p_{1}(n)=\sum _{k=1}^{n-1} k \left(\frac{1}{2 k+2}+\frac{1}{2 k+1}\right)\\ =\frac{1}{4} \left(4 n-2 \psi ^{(0)}(n+1)-\psi ^{(0)}\left(n+\frac{1}{2}\right)-2 \gamma -2+\psi ^{(0)}\left(\frac{3}{2}\right)\right)\\ =\frac{1}{4} \left(4 n-2 H_{n}-H_{n-\frac{1}{2}} -2+H_{\frac{1}{2}}\right)\tag{3c}$

$p_{2}(n)=\sum _{k=1}^n \frac{H_k}{2 k+1}\tag{3d}$

And I am stuck at evaluating $p_{2}(n)$.

Finite sums of products of harmonic numbers like $\sum_{k=1}^n H_{k} H_{2k}$

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