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Difference of harmonic number

Let \mathcal{H}_n^{(s)} denote the n – th harmonic number of weight s. Prove that

    \[\sum_{n=1}^{\infty}\frac{\mathcal{H}_n^2- \mathcal{H}_n^{(2)}}{(n+1)(n+2)}=2\]

Solution

Recall the generating function

(1)   \begin{equation*}  \sum_{n=1}^{\infty}\left(\mathcal{H}_n^2-\mathcal{H}_n^{(2)}\right)x^n = \frac{\ln^2(1-x)}{1-x} \end{equation*}

Thus (1) from 0 to t we get that

(2)   \begin{equation*} \sum_{n=1}^{\infty}\left(\mathcal{H}_n^2-\mathcal{H}_n^{(2)}\right)\frac{t^{n+1}}{n+1}=-\frac{1}{3}\ln^3(1-t)  \end{equation*}

Integrating (2) from 0 to 1 the result follows.

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