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removed the generalization - which would be better as a separate question; see John Ma's comment on this: https://math.meta.stackexchange.com/questions/19042/requests-for-reopen-undeletion-votes-etc-volume-01-2015-current-versio/28017#comment114319_28017
Martin Sleziak
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# Calculating the summation$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$

I need to find explicitly the following summation

$$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}, \quad H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$

From Mathematica, I checked that the answer is $$2$$. The same result is returned by WolframAlpha

This series can be found as Problem 3.59 (a) in the book Ovidiu Furdui: Limits, Series, and Fractional Part Integrals. Problems in Mathematical Analysis, Springer, 2013, Problem Books in Mathematics; where it is stated in the form

$$\sum_{k=1}^\infty\left(1+\frac12+\frac13+\dots+\frac1{n+1}\right)\frac1{n(n+1)}=2.$$

Some related thoughts:

Young
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