Somewhat more generally, Maple 18 says that

$$ \sum_{k=1}^n \dfrac{(-1)^k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} = -{\frac {\Gamma \left( v/2+2 \right) \Gamma \left( n-v/2 \right) }{ \left( v+2 \right) \Gamma \left( 1-v/2 \right) \Gamma \left( 1+v/2+ n \right) }} $$ which for $v=1$ gives you the desired identity.

It appears that Maple is using Zeilberger's algorithm.

Somewhat more generally, Maple 18 says that

$$ \sum_{k=1}^n \dfrac{(-1)^k k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} = -{\frac {\Gamma \left( v/2+2 \right) \Gamma \left( n-v/2 \right) }{ \left( v+2 \right) \Gamma \left( 1-v/2 \right) \Gamma \left( 1+v/2+ n \right) }} $$ which for $v=1$ gives you the desired identity.

It appears that Maple is using Zeilberger's algorithm.

Somewhat more generally, Maple 18 says that

$$ \sum_{k=1}^n \dfrac{(-1)^k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} = -{\frac {\Gamma \left( v/2+2 \right) \Gamma \left( n-v/2 \right) }{ \left( v+2 \right) \Gamma \left( 1-v/2 \right) \Gamma \left( 1+v/2+ n \right) }} $$ which for $v=1$ gives you the desired identity.

Somewhat more generally, Maple 18 says that

$$ \sum_{k=1}^n \dfrac{(-1)^k}{(n+k)(2k+v)} {n \choose k} {{n+k} \choose k} = -{\frac {\Gamma \left( v/2+2 \right) \Gamma \left( n-v/2 \right) }{ \left( v+2 \right) \Gamma \left( 1-v/2 \right) \Gamma \left( 1+v/2+ n \right) }} $$ which for $v=1$ gives you the desired identity.

It appears that Maple is using Zeilberger's algorithm.