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.
 A: 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:
$$ \sum_{k=1}^{n}\sum_{j=1}^{k}f(j)\cdot f(k)=\sum_{1\leq j\leq k\leq n}f(j)\cdot f(k) = \frac{1}{2}\left[\left(\sum_{k=1}^{n}f(k)\right)^2+\sum_{k=1}^{n}f(k)^2\right].$$
A: 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 $$H_{n}=H_{n+1}-\frac{1}{n+1}.$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\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}
