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

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.

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

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).

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}

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 5 '19 at 7:16
• @Ali Shather Indeed you are absolutely right. Thanks and have corrected it now. – omegadot Jul 5 '19 at 7:48
• looks good now :) – Ali Shadhar Jul 5 '19 at 7:50