How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{3}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}$$
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{n+1}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}$$
so
$$2I_n=\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)+\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)\left(\dfrac{1}{3}+\dfrac{1}{n+1}\right)+\cdots+\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)$$
Maybe this try is not usefull, so I think use other methods to solve it .
Thank you very much!
 A: Maybe I have got the definitive trick. I recall my previous $(5)$:
$$I_n=\frac{3}{2}\sum_{t=3}^{n+3}\frac{H_{t-1}^2-H_{t-1}^{(2)}}{t}.\tag{5}$$
Partial summation gives (I set $H_{0}^{(j)}=0$ for consistency):
$$\sum_{n=1}^{m}\frac{H_{n-1}^{(2)}}{n}=H_m H_{m-1}^{(2)}-\sum_{n=1}^{m-1}\frac{H_n}{n^2}=H_m H_m^{(2)}-\sum_{n=1}^m\frac{H_n}{n^2}=H_m H_m^{(2)}-H_{m}^{(3)}-\sum_{n=1}^{m}\frac{H_{n-1}}{n^2}.\tag{7}$$
This leads to:
$$I_n=\frac{3}{2}\sum_{t=1}^{n+3}\left(\frac{H_{t-1}^2}{t}+\frac{H_{t-1}}{t^2}\right)+\frac{3}{2}H_{n+3}^{(3)}-\frac{3}{2}H_{n+3}H_{n+3}^{(2)}.\tag{8}$$
Now we have (since $a^3-b^3=(a-b)(a^2+ab+b^2)$):
$$H_{j+1}^3-H_{j}^3 = \frac{H_{j+1}^2+H_{j+1} H_j +H_{j}^2}{j+1}=3\frac{H_j^2}{j+1}+3\frac{H_j}{(j+1)^2}+\frac{1}{(j+1)^3}.\tag{9}$$
Summing both sides of $(9)$ with $j$ that runs from $0$ to $n+2$ we have:
$$H_{n+3}^3-H_{n+3}^{(3)}=3\sum_{t=1}^{n+3}\left(\frac{H_{t-1}^2}{t}+\frac{H_{t-1}}{t^2}\right).\tag{10}$$
(An alternative proof of $(10)$, always based on partial summation, is given below by Matt Groff) 
If now we simply plug $(10)$ into $(8)$ we end with:
$$I_n = \frac{1}{2}H_{n+3}^3+H_{n+3}^{(3)}-\frac{3}{2}H_{n+3}H_{n+3}^{(2)},\tag{11}$$
that is way nicer than my previous $(6)$ and perfectly answers the question.
It is worth mentioning that, due to Greg Martin proof, this gives a closed expression for the sum
$$\sum_{i+j+k\leq n}\frac{1}{ijk}$$
and for the coefficients of the Taylor series of $\log^4(1-x)$ around zero.
Many many thanks to Greg Martin and Matt Groff for this brilliant piece of cooperative mathematics - should we ask to split the bounty in three?
A: Not a full answer, but hopefully helpful progress: rewriting
$$I_n=\sum_{k=1}^{n+1}\frac{H_{k}H_{n+2-k}}{k+1},$$
we recognize that
$$
\sum_{n=0}^\infty I_n x^{n+2} = \bigg( \sum_{k=1}^\infty \frac{H_k}{k+1} x^k \bigg) \bigg( \sum_{k=1}^\infty H_k x^k \bigg).
$$
Since $\sum_{j=1}^\infty \frac1j x^j = -\log(1-x)$ and $H_k = \sum_{j=1}^k \frac1j$, we see that
$$
\sum_{k=1}^\infty H_k x^k = \frac{-\log(1-x)}{1-x}.
$$
Integrating (and forcing the constant term to equal $0$) then gives
$$
\sum_{k=1}^\infty \frac{H_k}{k+1} x^{k+1} = \frac{\log^2(1-x)}2.
$$
Therefore
$$
\sum_{n=0}^\infty I_n x^n = \frac1{x^2} \bigg( \frac1x\frac{\log^2(1-x)}2 \bigg) \bigg( \frac{-\log(1-x)}{1-x} \bigg) = \frac{-\log^3(1-x)}{2x^3(1-x)}.
$$
We can use this generating function to try to get information about $\{I_n\}$. For example, if we define $\{c_n\}$ by
$$
-\log^3(1-x) = \sum_{n=1}^\infty c_n x^n,
$$
then we conclude that
$$
I_n = \frac12 \sum_{k=1}^{n+3} c_k.
$$
Note that
$$
c_n = \sum_{i+j+k=n} \frac1{ijk},
$$
so that
$$
I_n = \frac12 \sum_{i+j+k\le n+3} \frac1{ijk}.
$$
A: Continuing using both Greg Martin and Jack D'Aurizio's answers, I start with:
$$
\begin{align}
\sum_{k=1}^{m+1}\frac{(H_{k-1})^2}{k} &= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{(H_k+H_{k-1} )H_k}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \left( (H_{k-1}+\frac{1}{k}) + H_{k-1} \right)(H_{k-1}+\frac{1}{k})}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{ \left( 2H_{k-1}+\frac{1}{k}\right)(H_{k-1}+\frac{1}{k})}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{  2(H_{k-1})^2+\frac{3}{k}H_{k-1}+\frac{1}{k^2}}{k} } \\
3\sum_{k=1}^{m+1}\frac{(H_{k-1})^2}{k} &= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \frac{  \frac{3}{k}H_{k-1}+\frac{1}{k^2}}{k} } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \left( \frac{3}{k^2}H_{k-1}+\frac{1}{k^3} \right) } \\
&= (H_m)^2 H_{m+1} - \sum_{k=1}^m{ \left( \frac{3}{k^2}H_{k-1} \right) } + H_m^{(3)}
\end{align}
$$
