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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.

A thought from afar: $$ \begin{align} \frac{H_{n+1}}{n(n+1)} &=H_{n+1}\left(\frac1n-\frac1{n+1}\right)\tag1\\ &=\color{#C00}{H_{n+1}}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag2\\ &=\color{#C00}{\frac1{n+1}}\color{#090}{\frac1n}+\color{#C00}{H_n}\color{#090}{(H_n-H_{n-1})}-H_{n+1}(H_{n+1}-H_n)\tag3\\ &=\frac1n-\frac1{n+1}+\frac{H_n}{n}-\frac{H_{n+1}}{n+1}\tag4\\ &=\frac{H_n+1}{n}-\frac{H_{n+1}+1}{n+1}\tag5 \end{align} $$ What was done:
$(1)$: partial fractions
$(2)$: used $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(3)$: used $\color{#C00}{H_{n+1}=\frac1{n+1}+H_n}$ and $\color{#090}{\frac1n=H_n-H_{n-1}}$
$(4)$: partial fractions and $\frac1n=H_n-H_{n-1}$ and $\frac1{n+1}=H_{n+1}-H_n$
$(5)$: combined terms

Now this looks like Clement C's hint.

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(Big) hint:

$$\frac{H_{n+1}}{n(n+1)} = \frac{H_{n+1}}{n}-\frac{H_{n+1}}{n+1} = \frac{1}{n(n+1)} + \frac{H_{n}}{n}-\frac{H_{n+1}}{n+1}$$ so that you can get a telescopic series: for any $N\geq 1$, $$ \sum_{n=1}^N \frac{H_{n+1}}{n(n+1)} = \sum_{n=1}^N \frac{1}{n(n+1)} + \sum_{n=1}^N \frac{H_{n}}{n} - \sum_{n=2}^{N +1}\frac{H_{n}}{n} $$

Spoiler:

The value you should arrive at is $2$, which is $\sum_{n=1}^\infty \frac{1}{n(n+1)} + 1$.

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Hint: $$ \begin{align} \sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)} &=\sum_{n=1}^\infty\sum_{k=1}^{n+1}\frac1k\left(\frac1n-\frac1{n+1}\right)\\ &=\sum_{n=1}^\infty\frac1{n+1}\left(\frac1n-\frac1{n+1}\right) +\sum_{n=1}^\infty\sum_{k=1}^n\frac1k\left(\frac1n-\frac1{n+1}\right)\\ &=\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}-\frac1{(n+1)^2}\right) +\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1k\left(\frac1n-\frac1{n+1}\right) \end{align} $$ Now its simply summing telescoping series (three times).

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

divide both sides by $-n$ then take the sum, \begin{align} S&=\sum_{n=1}^\infty\frac{H_{n+1}}{n(n+1)}=-\int_0^1\ln(1-x)\left(\sum_{n=1}^\infty\frac{x^n}{n}\right)\ dx\\ &=\int_0^1\ln^2(1-x)\ dx=\int_0^1\ln^2x\ dx=2 \end{align}

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Noting that $$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1},$$ the sum may be rewritten as $$\sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} = \sum_{n = 1}^\infty \left (\frac{H_{n + 1}}{n} - \frac{H_{n + 1}}{n + 1} \right ).\tag1 \label1$$

Now as the harmonic numbers satisfy the recurrence relation $$H_{n + 1} = H_n + \frac{1}{n + 1},$$ we have \begin{align} \sum_{n = 1}^\infty \frac{H_{n+1}}{n(n + 1)} &= \sum_{n = 1}^\infty \left [\frac{H_n}{n(n+1)} + \frac{1}{n(n +1)} - \frac{1}{(n + 1)^2} \right ]\\ &= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n + 1)} - \underbrace{\sum_{n = 1}^\infty \frac{1}{(n + 1)^2}}_{n \, \mapsto n - 1}\\ &= \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} + \sum_{n = 1}^\infty \frac{1}{n(n+1)} - \sum_{n = 1}^\infty \frac{1}{n^2} + 1\tag2\\ &= \zeta (2) + 1 - \zeta (2) + 1\\ &= 2 \end{align} In (2) an evaluation for the first sum can be found here while the second sum telescopes and has a sum equal to one.

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  • $\begingroup$ just a little note. $\sum_{n=2}^\infty\frac{H_n}{n} $ is divergent so we can not cancel these two sums. so we better simplify before distributing. $\endgroup$ – Ali Shadhar Jul 5 '19 at 7:16
  • $\begingroup$ @Ali Shather Indeed you are absolutely right. Thanks and have corrected it now. $\endgroup$ – omegadot Jul 5 '19 at 7:48
  • $\begingroup$ looks good now :) $\endgroup$ – Ali Shadhar Jul 5 '19 at 7:50

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