The harmonic numbers are given by $$H\_n=\sum\_{k=1}^n\frac{1}{k}.$$
Numerical calculation suggests
$$
\sum\_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum\_{i=1}^{k}\frac{1}{n+i}=(-1)^nH\_n.
$$
I can not give a proof of this identity. How to prove it? 


Hints, references or proof are all welcome.



A combinatorial identity involving harmonic numbers
First we prove the formula
$$\sum\_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k} = (-1)^n\binom xn,\tag{1}$$
which is special case of Vandermonde's theorem:
$$\begin{aligned}
\sum\_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k} 
 &= \sum\_{k=0}^n \binom n{n-k} \binom{-x-1}{k}\\
 &= \binom{n-x-1}{n} = (-1)^n\binom xn.
\end{aligned}$$
Since $(1)$ is an identity of polynomials in $x$, we may differentiate it with respect to $x$, obtaining 
$$\sum\_{k=0}^n (-1)^k\binom{n}{k}\binom {x+k}{k}\sum\_{i=1}^k\frac{1}{x+i}
 =(-1)^n\binom xn \sum\_{i=0}^{n-1}\frac{1}{x-i}.
$$
The OP's identity follows on setting $x=n$.


The same method can be used to prove many identities involving harmonic numbers.


