<s>[INST] What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$? 
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.
 [/INST] We seek to find a closed form of
$$\sum_{r=0}^n (-1)^r {n\choose r}^{-1}.$$
Recall from MSE
4316307  the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$1+ \sum_{r=0}^n (-1)^r r [z^n] \log\frac{1}{1-z} (z-1)^{n-r}
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n
[v^n] \frac{1}{1-v}
\sum_{r\ge 0} (-1)^r r (z-1)^{-r} v^r
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^n
[v^n] \frac{1}{v-1} \frac{v/(z-1)}{(1+v/(z-1))^2}
\\ = 1+ [z^n] \log\frac{1}{1-z} (z-1)^{n+1}
[v^{n-1}] \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$
The contribution from $v$ is
$$\;\underset{v}{\mathrm{res}}\;
\frac{1}{v^n} \frac{1}{v-1} \frac{1}{(z-1+v)^2}.$$
The residue at infinity is zero so we may compute this by taking minus
the residue at one and minus the residue at $v=1-z.$ (Residues sum to
zero.) We get for the  former
$$- [z^{n+2}] \log\frac{1}{1-z} (z-1)^{n+1}
= - {n+2\choose 1}^{-1} = -\frac{1}{n+2}$$
and the latter
$$- \left.\left( \frac{1}{v^n} \frac{1}{v-1} \right)'\right|_{v=1-z}
= - \left.\left( -\frac{n}{v^{n+1}} \frac{1}{v-1}
- \frac{1}{v^n} \frac{1}{(v-1)^2} \right)\right|_{v=1-z}
\\ = - \frac{n}{z (1-z)^{n+1}} + \frac{1}{(1-z)^n z^2}.$$
We get for the first term
$$- (-1)^{n+1} n [z^{n+1}] \log\frac{1}{1-z}
= (-1)^n \frac{n}{n+1}.$$
and for the second
$$(-1)^n [z^{n+2}] \log\frac{1}{1-z} (z-1)
= (-1)^n \frac{1}{n+1} - (-1)^n \frac{1}{n+2}.$$
Collecting everything we find
$$1+(-1)^n - \frac{1}{n+2} (1+(-1)^n).$$
This will produce zero when $n$ is odd. For $n$ even we find
$$\bbox[5px,border:2px solid #00A000]{
2 \frac{n+1}{n+2}.}$$
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