# Find a closed form to $\sum\limits_{i=2}^{n} \frac{H_i}{i+1}$

So I'm trying to do this annoying proof and without going into further details I think after quite a while of thinking I found it. Now I get stuck with an annoying sum (of sums..) where I don't quite know if there exists a closed form and if so how to find it.

So as I already said I try to find a closed form to the following series $\sum\limits_{i=2}^{n} \frac{H_i}{i+1}$ where $H_n$ is the harmonic series ($\sum\limits_{i=1}^{n} \frac{1}{i}$).

So yeah any hint for a closed form of the above series is more than welcome! Thanks in advance for any help.

• The authors in the following wiki try to give a proof for the closed form of $H_n$ which I would mistrust due to the slightly dodgy way of representing an integral representation of an alternating sum. Check it out though, may get you somewhere en.wikipedia.org/wiki/Harmonic_number#Calculation – complexmanifold Oct 12 '16 at 13:10
• Yeah well besides the dodgy integral this doesn't really help me in this case. Since it's not only the harmonic series but there's also a sum and a division.. Thanks though! – Desperate Oct 12 '16 at 13:19
• I know, but I thought that with a little algebra you could import the division by $i+1$ and manipulate. – complexmanifold Oct 12 '16 at 13:22
• Well I kind of have a closed form if it weren't for the division .. $\sum\limits_{i=1}^n H_i = (n+1) \cdot (H_{n+1} - 1)$ .. what I want to say is that it doesn't matter if I have the harmonic series still in there as long as I can get the other sum away. – Desperate Oct 12 '16 at 13:49

A preliminary manipulation:

$$\sum_{i=2}^{n}\frac{H_i}{i+1}=-\frac{1}{2}+\sum_{i=1}^{n}\frac{H_i}{i+1}=-\frac{1}{2}+\sum_{i=1}^{n}\frac{H_{i+1}}{i+1}-\sum_{i=1}^{n}\frac{1}{(i+1)^2}$$ gives: $$\sum_{i=2}^{n}\frac{H_i}{i+1}=\sum_{k=1}^{n+1}\frac{H_k}{k}-\left(H_{n+1}^{(2)}+\frac{1}{2}\right)\tag{1}$$ and now: $$\sum_{k=1}^{n+1}\frac{H_k}{k}=\sum_{k=1}^{n+1}\frac{1}{k}\sum_{j=1}^{k}\frac{1}{j}=\sum_{1\leq j\leq k\leq n+1}\frac{1}{j\cdot k}=\frac{1}{2}\left[\left(\sum_{i=1}^{n+1}\frac{1}{i}\right)^2+\sum_{i=1}^{n+1}\frac{1}{i^2}\right] \tag{2}$$ leads to: $$\sum_{i=2}^{n}\frac{H_i}{i+1}=\color{red}{\frac{H_{n+1}^2-H_{n+1}^{(2)}-1}{2}}.\tag{3}$$ Here $H_m^{(2)}$ stands for $\sum_{k=1}^{m}\frac{1}{k^2}$, as usual. We exploited $H_{m}=H_{m+1}-\frac{1}{m+1}$.

Non-believers may just take $(3)$ as a claim and prove it through induction on $n$.
Anyway, the crucial part $(2)$ is just an instance of the following identity:

$$\sum_{k=1}^{n}\sum_{j=1}^{k}f(j)\cdot f(k)=\sum_{1\leq j\leq k\leq n}f(j)\cdot f(k) = \frac{1}{2}\left[\left(\sum_{k=1}^{n}f(k)\right)^2+\sum_{k=1}^{n}f(k)^2\right].$$

• +1 IIRC, there is also an expression involving Stirling numbers of the second kind. Probably not what is expected here though. – Jean-Claude Arbaut Oct 12 '16 at 17:10
• Stirling numbers of the first kind in fact. :) – Jean-Claude Arbaut Oct 12 '16 at 20:00
• @Jean-ClaudeArbaut: clearly so, just written in a explicit way, depending on generalized harmonic numbers. – Jack D'Aurizio Oct 12 '16 at 20:01
• Well first off thanks a bunch! I agree that Stirling numbers are way beyond what I need thanks for that aswell though! – Desperate Oct 12 '16 at 20:12
• aaaaaah I see!!! sorry for the inconvenience I was thinking it would be way less obvious – Desperate Oct 12 '16 at 20:29

It can be proven also using summation by parts $$S=\sum_{i=2}^{n}\frac{H_{i}}{i+1}=\sum_{i=2}^{n}\frac{1}{i+1}\cdot H_{i}=H_{n}\left(H_{n+1}-\frac{3}{2}\right)-\sum_{i=2}^{n-1}\frac{\left(H_{i+1}-\frac{3}{2}\right)}{i+1}$$ hence $$\sum_{i=2}^{n}\frac{H_{i}}{i+1}=H_{n}\left(H_{n+1}-\frac{3}{2}\right)-\sum_{i=2}^{n-1}\frac{H_{i+1}}{i+1}+\frac{3}{2}\sum_{i=2}^{n-1}\frac{1}{i+1}$$ $$=H_{n}H_{n+1}-\sum_{i=2}^{n}\frac{H_{i}}{i+1}+\frac{H_{n}}{n+1}-\sum_{i=2}^{n-1}\frac{1}{\left(i+1\right)^{2}}-\frac{9}{4}$$ so $$\sum_{i=2}^{n}\frac{H_{i}}{i+1}=\color{red}{\frac{1}{2}\left(H_{n}H_{n+1}+\frac{H_{n}}{n+1}-H_{n}^{\left(2\right)}-1\right)}.$$ Note that this is the same result of the other answers, since $$H_{n}=H_{n+1}-\frac{1}{n+1}.$$

• (+1) I was going to add this approach to my previous answer, but there is no need now. Well done, as always. – Jack D'Aurizio Oct 13 '16 at 13:45
