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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.

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  • $\begingroup$ 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 $\endgroup$ – complexmanifold Oct 12 '16 at 13:10
  • $\begingroup$ 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! $\endgroup$ – Desperate Oct 12 '16 at 13:19
  • $\begingroup$ I know, but I thought that with a little algebra you could import the division by $i+1$ and manipulate. $\endgroup$ – complexmanifold Oct 12 '16 at 13:22
  • $\begingroup$ 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. $\endgroup$ – Desperate Oct 12 '16 at 13:49
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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:

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

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

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    $\begingroup$ (+1) I was going to add this approach to my previous answer, but there is no need now. Well done, as always. $\endgroup$ – Jack D'Aurizio Oct 13 '16 at 13:45
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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = 2}^{n}{H_{i} \over i + 1} & = \sum_{i = 2}^{n}\braces{{1 \over 2}\bracks{\pars{H_{i} + {1 \over i + 1}}^{2} - H_{i}^{2} - \pars{1 \over i + 1}^{2}}} \\[5mm] & = {1 \over 2}\sum_{i = 2}^{n}H_{i + 1}^{2} - {1 \over 2}\sum_{i = 2}^{n}H_{i}^{2} - {1 \over 2}\sum_{i = 2}^{n}{1 \over \pars{i + 1}^{2}} \\[5mm] & = {1 \over 2}\sum_{i = 3}^{n + 1}H_{i}^{2} - \bracks{{1 \over 2}\,H_{2}^{2} + {1 \over 2}\sum_{i = 3}^{n + 1}H_{i}^{2} - {1 \over 2}\,H_{n + 1}^{2}} - {1 \over 2}\sum_{i = 3}^{n + 1}{1 \over i^{2}}\quad \pars{~\mbox{note that}\ {1 \over 2}\,H_{2}^{2} = {9 \over 8}~} \\[5mm] & = {1 \over 2}\,H_{n + 1}^{2} - {9 \over 8} - \pars{-{1 \over 2} - {1 \over 8} + {1 \over 2}\sum_{i = 1}^{n + 1}{1 \over i^{2}}} = \bbox[8px,border:1px groove navy]{{1 \over 2}\,H_{n + 1}^{2} - {1 \over 2} - {1 \over 2}\,H_{n + 1}^{\pars{2}}} \end{align}

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