TITLE: How to compute the sum $\sum_{r=0}^n \frac{(-1)^r}{{n \choose r}}$
QUESTION [5 upvotes]: 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

REPLY [5 votes]: 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*}