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I am interested in the proof of $$\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2, \quad H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}$$

This result can be verified by Mathematica or 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:

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  • $\begingroup$ Have you tried the integral representation of harmonic numbers? $\endgroup$ – Yuriy S Mar 11 '18 at 2:10
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    $\begingroup$ Thank you for doing this, Martin -- I presume you'll mention this in the relevant Meta thread? $\endgroup$ – pjs36 Mar 11 '18 at 2:15
  • $\begingroup$ Summation by parts looks very natural here---the differences of the $H_n$'s are clear, as are the partial sums of [\frac{1}{n(n+1)=\frac{1}{n} - \frac{1}{n+1}] $\endgroup$ – Jamie Radcliffe Mar 11 '18 at 2:16
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Well.

Since

\begin{align} \frac{1}{(n+1)n}= \frac{1}{n}-\frac{1}{n+1} \end{align} then we see that \begin{align} \frac{H_{n+1}}{n(n+1)} =&\ \frac{H_{n+1}}{n} - \frac{H_{n+1}}{n+1}\\ =&\ \frac{H_n}{n}-\frac{H_{n+1}}{n+1}+\frac{1}{n(n+1)}. \end{align} Hence we have that \begin{align} \sum^\infty_{n=1}\frac{H_{n+1}}{n(n+1)} = \sum^\infty_{n=1}\left(\frac{H_n}{n}-\frac{H_{n+1}}{n+1} \right)+\sum^\infty_{n=1}\left(\frac{1}{n}-\frac{1}{n+1} \right) = H_1 + 1 = 2. \end{align}

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Through generating functions: $$ -\log(1-x)=\sum_{n\geq 1}\frac{x^n}{n},\qquad \frac{-\log(1-x)}{1-x}=\sum_{n\geq 1} H_{n} x^n\tag{1} $$ for any $x\in(0,1)$. By considering that $\int_{0}^{1}-\log(x)\,dx=1$ and that $\int_{0}^{1}x^n(1-x)\,dx=\frac{1}{(n+1)(n+2)}$ $$ -\log(1-x)=\sum_{n\geq 1}H_n x^n(1-x),\qquad 1=\sum_{n\geq 1}\frac{H_{n}}{(n+1)(n+2)}.\tag{2} $$ Since $$ \sum_{n\geq 1}\frac{1}{(n+1)^2(n+2)}+\frac{1}{(n+1)(n+2)^2}=\sum_{n\geq 1}\frac{1}{(n+1)^2}-\frac{1}{(n+2)^2}=\frac{1}{4}\tag{3}$$ we get $$ \frac{5}{4}=\sum_{n\geq 1}\frac{H_{n+2}}{(n+1)(n+2)},\qquad \sum_{n\geq 1}\frac{H_{n+1}}{n(n+1)}=\frac{5}{4}+\frac{H_2}{2}=2.\tag{4}$$

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For this solution, we'll make use of the telescoping sum

$$\sum_{n=1}^\infty{1\over n(n+1)}=\sum_{n=1}^\infty\left({1\over n}-{1\over n+1} \right)=1$$

and freely interchange sums, derivatives, and integrals.

First, let

$$H_{n+1}(x)=x+{x^2\over2}+\cdots+{x^{n+1}\over n+1}$$

so that

$$H_{n+1}'(x)=1+x+\cdots+x^n={1-x^{n+1}\over1-x}$$

Now let

$$F(x)=\sum_{n=1}^\infty{H_{n+1}(x)\over n(n+1)}$$

so that

$$\begin{align} F'(x)&=\sum_{n=1}^\infty{H_{n+1}'(x)\over n(n+1)}\\ &={1\over1-x}\left(\sum_{n=1}^\infty{1\over n(n+1)}-\sum_{n=1}^\infty{x^{n+1}\over n(n+1)} \right)\\ &={1\over1-x}\left(1-\sum_{n=1}^\infty\left({x^{n+1}\over n}-{x^{n+1}\over n+1} \right)\right)\\ &={1\over1-x}\left(1+\sum_{n=2}^\infty{x^n\over n}-x\sum_{n=1}^\infty{x^n\over n} \right)\\ &={1\over1-x}\left(1-x+\sum_{n=1}^\infty{x^n\over n}-x\sum_{n=1}^\infty{x^n\over n} \right)\\ &=1+\sum_{n=1}^\infty{x^n\over n} \end{align}$$

and thus, since $F(0)=0$, we have

$$\sum_{n=1}^\infty{H_{n+1}\over n(n+1)}=F(1)=\int_0^1F'(x)dx=1+\sum_{n=1}^\infty{1\over n(n+1)}=1+1=2$$

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using the fact that $\displaystyle -\int_0^1x^n\ln(1-x)\ dx=\frac{H_{n+1}}{n+1}$

divide both sides by $n$ then take the sum, we get

$$\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{x^n}{n}\ dx=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2$$

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$$\sum_{n=1}^N \frac{H_{n+1}}{n(n+1)} =\sum_{n=1}^N H_{n+1}\left(\frac1n-\frac1{n+1}\right) =H_2-\frac{H_{N+1}}{N+1}+\sum_{n=2}^{N}\frac{H_{n+1}-H_n}{n} =H_2-\frac{H_{N+1}}{N+1}+\sum_{n=2}^{N}\frac1{n(n+1)} \to\frac32+\sum_{n=2}^\infty\frac1{n(n+1)}$$ as $N\to\infty$ etc.

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