Calculating the summation $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$

Question

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.

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and $H_{n+2}-H_{n+1}=\frac{1}{n+2}$, hence summation by parts converts such series into an elementary (telescopic) one. — Jack D'Aurizio, Feb 25 '18 at 16:47
A question about the same series: How to show $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}=2$? — Martin Sleziak, May 15 '19 at 06:28
Since the question was undeleted, perhaps it is worth adding also links to two posts on meta about this question: https://math.meta.stackexchange.com/questions/19042/requests-for-reopen-undeletion-votes-etc-volume-01-2015-07-2018/28017#28017 and https://math.meta.stackexchange.com/questions/28692/requests-for-reopen-undeletion-votes-volume-07-2018-today/30181#30181 — Martin Sleziak, May 15 '19 at 07:33

score 4 · Accepted Answer · answered Feb 25 '18 at 04:08

(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 $\sum_{n=1}^\infty \frac{1}{n(n+1)} + 1$.

score 4 · Answer 2 · answered Feb 25 '18 at 04:33

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).

omegadot · Answer 3 · 2019-07-05T07:46:50.563

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.

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. — Ali Shadhar, Jul 05 '19 at 07:16
@Ali Shather Indeed you are absolutely right. Thanks and have corrected it now. — omegadot, Jul 05 '19 at 07:48

score 2 · Answer 4 · answered Jul 05 '19 at 07:09

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}

Calculating the summation $\sum_{n=1}^{\infty}\frac{H_{n+1}}{n(n+1)}$

4 Answers4