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As an extension to my question Infinite sum of harmonic number over polynomial of 2nd degreee I found a general formula for a polynomial of arbitrary degree which I had presented as a self answer but find it interesting enough to move it here as a separate post. This formula covers many special cases discussed here and elsewhere.

Question: prove the formula.
Hint: make a partial fraction decomposition and then see the answer of Jack d'Aurizio to the reference above.

The formula

Let $H_k = \sum_{i=1}^k 1/i$ be the harmonic number and $p_1,\dotsc,p_n$ the (possibly complex) pairwise distinct (negative) roots of a polynomial of degree $n$ (for multiple roots see below).

Define

$$s(p_1,\dotsc,p_n)=\sum _{k=1}^{\infty } \frac{H_k}{\prod _{i=1}^n (p_i +k)}\tag{1}$$

and

$$f(p_1,\dotsc,p_n)=\frac{1}{2} (-1)^n \sum _{i=1}^n \frac{h(p_i)}{\prod _{j=1,j\neq i}^n (p_i-p_j)}\tag{2}$$

where

$$h(a)=H_{a-1}^2-\psi ^{(1)}(a)+\zeta (2)\tag{3}$$

Then

$$s(p_1,\dotsc,p_n)=f(p_1,\dotsc,p_n)\tag{4}$$

In the case of originally multiple roots we start from pairwise distinct roots (if necessary "artificially" made distinct) and take appropriate limits of $f(p)$. This means effectively to apply the rule of l'Hospital which generates derivatives of $h(a)$.

Possibly useful formulas for transforming the final results are

$$\psi ^{(0)}(x+1)=H_x^{(1)}-\gamma\tag{5a}$$

$$\psi ^{(k)}(x+1)=(-1)^{k+1} k! \left(\zeta (k+1)-H_x^{(k+1)}\right)\tag{5b}$$

Where $H_x^{(m)}$ is the generalized harmonic number.

Examples

Many specific formulas can be written down using the relation established here.

Example 1

As a non trivial check and an application of multiple roots I rederived the Borwein-Borwein formulas (19) - (21) given in http://mathworld.wolfram.com/HarmonicNumber.html for sums of the type

$$\sum_{k=1}^\infty \frac{H_k}{k^n}$$

Related: Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

Example 2

This is the "initiating" example of How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$

$$\sum_{k=1}^\infty \frac{H_k}{(2k+1)(2k+2)}$$

This can be written and calculated as

$$\frac{1}{4} f\left(\left\{\frac{1}{2},1\right\}\right) = \frac{\pi ^2}{12}-\frac{1}{4} H_{-\frac{1}{2}}^2 = \frac{\pi ^2}{12}- \log(2)^2$$

where we have used that

$$H_{-\frac{1}{2}} = -\log(4)$$

Example 3

Purely imaginary roots. Setting $a=i$, $b=-i$ we obtain

$$\sum _{k=1}^{\infty } \frac{H_k}{k^2+1}=\frac{1}{4} \left(2 \gamma (1+\pi \coth (\pi ))+i \left(\psi ^{(0)}(-i)^2-\psi ^{(0)}(i)^2-\psi ^{(1)}(1-i)+\psi ^{(1)}(1+i)\right)\right)$$

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Jack D'Aurizio Oct 23 '17 at 14:21

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