The $n$-th harmonic number is defined as
$$
H\_n=\sum\_{k=1}^{n}\frac{1}{k},
$$
and the generalized harmonic numbers are defined by
$$
H\_{n}^{(r)}=\sum\_{k=1}^{n}\frac{1}{k^r}.
$$
Recently, I have found the following combinatorial identity involving the second-order harmonic numbers (I have computational evidence).


**Question:**
\begin{align}
\sum\_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s }{s+1}H\_{s}^{(2)}=\frac{2(-1)^m}{m+1}\sum\_{s=0}^m H\_{s}^{(2)}. 
\end{align}
Is this a known combinatorial identity? Any proof or reference?
However, if I replace $H\_s^{(2)}$ by other generalized harmonic numbers in the above identity, I can not find some similar identities.


**Note:**
This combinatorial identity was motivated by the following identity
\begin{align}
\sum\_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m.
\end{align}
One can refer to [How to prove $\sum\_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$?](https://math.stackexchange.com/questions/1559612/how-to-prove-sum-s-0m2s-choose-ss-choose-m-s-frac-1ss1-1m)


I appreciate any hints, pointers etc.!



A combinatorial identity involving generalized harmonic numbers
The identity
$$
\begin{align}
\sum\_{s=1}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s }{s+1}H\_{s}^{(2)}=\frac{2(-1)^m}{m+1}\sum\_{s=1}^m H\_{s}^{(2)}. \tag{1}
\end{align}
$$
is equivalent to the following identity
$$
\sum\_{s=1}^{m}{2s\choose s}\frac{H\_s^{(2)}}{s+1}(x-x^2)^s=\frac{2\text{Li}\_2(x)}{1-x}-\frac{\ln^2(1-x)}{x},\tag{2}
$$
where $\text{Li}\_2$ is dilogarithm function. To prove this, first note that
$$
\sum\_{s=1}^m H\_{s}^{(2)}=\sum\_{s=1}^m \sum\_{k=1}^s \frac{1}{k^2}=\sum\_{k=1}^m \frac{1}{k^2}\sum\_{s=k}^m 1=\\ \sum\_{k=1}^m \frac{1}{k^2}(m+1-k)=(m+1)H\_{m}^{(2)}-H\_{m}^{(1)}.
$$
Now using the generating functions (see [link1](https://en.wikipedia.org/wiki/Harmonic_number#Generating_functions), [link2](https://en.wikipedia.org/wiki/Harmonic_number#Generalization))
$$
\sum\_{m=1}^\infty \frac{H\_{m}^{(1)}}{m+1}x^m=\frac{\ln^2(1-x)}{2x},\qquad \sum\_{m=1}^\infty H\_{m}^{(2)}x^m=\frac{\text{Li}\_2(x)}{1-x}
$$
and proceeding as in this [answer](https://math.stackexchange.com/a/1560084/285751) one obtains
\begin{align}
\sum\_{s=1}^{\infty}\binom{2s}{s}\frac{(-1)^s}{s+1}x^s (1+x)^sH\_s^{(2)}=&\sum\_{m=1}^{\infty}\frac{2(-x)^m}{m+1}\left[(m+1)H\_{m}^{(2)}-H\_{m}^{(1)}\right]=\\
&\frac{2\text{Li}\_2(-x)}{1+x}-\frac{\ln^2(1+x)}{-x}
\end{align}
which is equivalent to $(2)$.


$\bf{Proof\ of\ eq.(2)}$
The [generating function](https://en.wikipedia.org/wiki/Catalan_number#First_proof) of Catalan numbers
$$
f(x)=\sum\_{n=0}^\infty {2n\choose n}\frac{x^n}{n+1}=\frac{1-\sqrt{1-4x}}{2x}.\tag{3}
$$
The generating function of ${2s\choose s}\frac{H\_s^{(2)}}{s+1}$ can be obtained from $(3)$ by integrating the identity
$$
\frac{f(x)-f(xy)}{1-y}=\sum\_{n=1}^\infty {2n\choose n}\frac{x^n}{n+1}(1+y+...+y^{n-1})
$$
as follows
\begin{align}
\sum\_{s=1}^{m}{2s\choose s}\frac{x^s}{s+1}H\_s^{(2)}=&\int\_0^1\frac{dt}{t}\int\_0^t \frac{f(x)-f(xy)}{1-y}dy=\\
&\int\_0^1\frac{dt}{t}\int\_0^t\frac{dy}{1-y}\left(\frac{1-\sqrt{1-4x}}{2x}-\frac{1-\sqrt{1-4xy}}{2xy}\right)=\\
&-\int\_0^1\frac{\ln t}{1-t}\left(\frac{1-\sqrt{1-4x}}{2x}-\frac{1-\sqrt{1-4xt}}{2xt}\right)dt
\end{align}
[the last line follows after integrating by parts with respect to $t$]. After substitution $\sqrt{1-4xt}=u$ this last integral becomes
$$
\frac{\pi^2}{6}\cdot \frac{1-\sqrt{1-4x}}{2x}+4\int\limits\_{\sqrt{1-4x}}^1\frac{\ln\frac{1-y^2}{4x}}{(1+y)(y^2-1+4x)}ydy.
$$
So after replacing $x$ with $x-x^2$ one obtains
$$
\sum\_{s=1}^{m}{2s\choose s}\frac{(x-x^2)^s}{s+1}H\_s^{(2)}=\frac{\pi^2}{6(1-x)}+4\int\limits\_{1-2x}^1\frac{\ln\frac{1-y^2}{4x(1-x)}}{(1+y)(y^2-(1-2x)^2)}ydy
$$
This last integral can be calculated by Mathematica in terms of dilogarithm functions, or manually after expanding $\frac{y}{(1+y)(y^2-(1-2x)^2)}$ into partial fractions. Then using functional equations for dilogarithm I verified that the resulting expression equals $\frac{2\text{Li}\_2(x)}{1-x}-\frac{\ln^2(1-x)}{x}$.


