I found in Matlab that $$ \sum_{k=0}^n~\frac{(-1)^k}{k+x}\binom{n}{k}\binom{n+k}{k}=0$$ for $x$ integer with $1\leq x< n$ only (I am about 95% sure of this since the sum is numerically unstable and cannot give accurate results for $n>\approx 20$)
Is there a way of obtaining a similar result for real $0<x<1$? Is analytic continuation useful?
If it is useful, for $x=0$ we have $$\sum_{k=1}^n~\frac{(-1)^k}{k}\binom{n}{k}\binom{n+k}{k}=-2\sum_{k=0}^n~\frac{1}{k}$$
From Mathematica and/or Wolfram Alpha, I get $$ \sum_{k=0}^n~\frac{(-1)^k}{k+x}\binom{n}{k}\binom{n+k}{k}=\frac{(x-1)! (n-x)!}{x (x+1) (-x-2)! (n+x)!}. $$ However, the intermediate steps are unavailable.
I guess one can obtain the same identity with Gosper's algorithm. One can find a comprehensive introduction of Gosper's algorithm in the book Concrete Mathematics.