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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.!

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.!

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 H_{s}^{(2)}}{s+1}=\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.!

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.!