The sum $\sum_{k=0}^{n} (-1)^k \frac{{n \choose k}}{{n+k \choose k}}=\frac{1}{2}$

Question

The sum $$\sum_{k=0}^{n} (-1)^k \frac{{n \choose k}}{{n+k \choose k}}=\frac{1}{2}?$$ can be checked by Mathematica. Here, the question is how to do it by hand?

Answer 1

Let $$S_n=\sum_{k=0}^{n} (-1)^k \frac{{n \choose k}}{{n+k \choose k}}= \frac{n!^2}{(2n)!} \sum_{k=0}^{n}(-1)^k \frac{(2n)!}{(n-k)!~ (n+k)!}=\frac{1}{{2n \choose n}} \sum_{k=0}^n (-1)^{n-k} \frac{(2n)!}{k! (2n-k)!}=(-1)^n \frac{1}{{2n \choose n}} \sum_{k=0}^{n} (-1)^k {2n \choose k}=\frac{1}{2}.$$