<s>[INST] 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!
 [/INST] 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
$$
I_n = \frac12 \sum_{i+j+k\le n+3} \frac1{ijk}.
$$
</s>