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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 $\sum_{s=0}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s}{s+1}=(-1)^m$?

I appreciate any hints, pointers etc.!

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    $\begingroup$ The WZ-method provides an algorithm for verifying any identities like this (assuming they're true) from a wide class. en.wikipedia.org/wiki/Wilf–Zeilberger_pair $\endgroup$ – Greg Martin Dec 26 '15 at 6:39
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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, link2) $$ \sum_{m=1}^\infty \frac{H_{m}^{(1)}}{m+1}x^m=\frac{\ln^2(1-x)}{2x},\qquad \sum_{m=1}^\infty H_{m}^{(2)}x^m=\frac{\text{Li}_2(x)}{1-x} $$ and proceeding as in this answer 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 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 $\frac{2\text{Li}_2(x)}{1-x}-\frac{\ln^2(1-x)}{x}$.

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