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!

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?
