How to compute the sum $\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}$

Question

Consider the sum
$$\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}.$$ I know the sum is zero when $n$ is odd (pretty simple). The sum is $2-\frac{2}{2 + n}$ when $n$ is even.

Can somebody provide a proof in the even case? Thanks

Answer 1

Start by writing ${1/ {n\choose r}}=(n+1)\int_0^1 u^r(1-u)^{n-r}\,du.$ Multiplying by $(-1)^r$ and adding over $r$ gives \begin{eqnarray*} \sum_{r=0}^n (-1)^r/{n\choose r} &=&(n+1)\int_0^1 \sum_{r=0}^n (-u)^r(1-u)^{n-r}\,du\\ &=&(n+1)\int_0^1 [(1-u)^{n+1}+(-1)^nu^{n+1}]\,du\\ &=&{n+1\over n+2}(1+(-1)^n). \end{eqnarray*}