TITLE: What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?
QUESTION [20 upvotes]: Find a closed form expression for $$\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}$$
where $n$ is an even positive integer.

I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, elementary solution, if possible.
Thanks.

REPLY [3 votes]: For a fixed $k$, if we regard $\frac{1}{\binom{n}{k}}$ as a meromorphic function of $n$, we have:
$$ \text{Res}\left(\binom{n}{k}^{-1},\,n=r\right)=k(-1)^{k+r+1}\binom{n-1}{r} $$
for every $r\in[0,k-1]$. That implies:
$$ \text{Res}\left(\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}},n=0\right)=0-1-2-\ldots-n=-\binom{n+1}{2}$$
as well as:
$$ \text{Res}\left(\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}},n=1\right)=0+2+6+12+\ldots+2\binom{n}{2}=2\binom{n+1}{3}$$
so, by induction:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}^{-1}=\sum_{h=0}^{n-1}\frac{(-1)^{h+1}(h+1)}{n-h}\binom{n+1}{h+2}=(n+1)\sum_{h=0}^{n-1}\frac{(-1)^{h+1}}{h+2}\binom{n}{h}$$
but the last sum is easily manageable through usual techniques, since it is clearly related with $\int_{0}^{1}x(1-x)^n\,dx$.