# 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.

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
• For gosh sakes, the asker has been suspended for their low quality posts this week, which only happens when an asker habitually asks poor questions, and yet we have three users, in addition to a mod (because that mod happened to answer the question, and doesn't want their answer deleted), who, nonetheless seek to undelete and reopen this question. – Namaste Mar 11 '18 at 1:02
• – user296602 Mar 11 '18 at 1:13
• @user296602 the posts content is crap. Thanks for making my point. – Namaste Mar 11 '18 at 1:16
• To whoever it may concern: please, delete the question once and for all, or stop downvoting its answers. It makes no sense to leave it open but retaliate on the answers. – Clement C. Mar 11 '18 at 1:16

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

• Why the downvote? – Clement C. Mar 5 '18 at 1:07
• It is to be expected since the question was just undeleted and there were some people who were quite against the undeletion. Don't worry, the downvotes don't reflect any on the answers, just that the question that they thought should be deleted was answered. I got one too. – robjohn Mar 7 '18 at 17:00
• Not true, @robjohn. The downvotes received reflect heavily on both the answers and the answerers. You may gain peace of mind by recreating a narrative that soothes your soul, but that doesn't make it true. Three downvotes here have a much better explanation than you're even able to contemplate, and you'd have more too, if you weren't flouting a diamond alongside your username, dear diamond robjohn. Just stop trying to appease others and yourself that answers to poor questions asked by users suspended for posting low quality questions are commendable. – Namaste Mar 11 '18 at 1:07
• @amWhy In this case (I don't care about my peace of mind, and am not attached to any narrative) what is wrong with my answer? I am fine with the question being deleted, I haven't cast any vote, yet my answer gets downvoted. In short, to whoever it may concern: delete the goddarn question once and for all, or stop downvoting its answers! – Clement C. Mar 11 '18 at 1:11

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