What is $\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}$?

Brute force with the Beta function: $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} &=\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+1)}\tag{1a}\\ &=(n+1)\sum_{k=0}^n(-1)^k\frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\tag{1b}\\ &=(n+1)\sum_{k=0}^n(-1)^k\operatorname{B}(k+1,n-k+1)\tag{1c}\\ &=(n+1)\int_0^1\sum_{k=0}^n(-1)^kt^k(1-t)^{n-k}\,\mathrm{d}t\tag{1d}\\ &=(n+1)\int_0^1(1-t)^n\frac{1+\left(\frac t{1-t}\right)^{n+1}}{1+\frac{t}{1-t}}\,\mathrm{d}t\tag{1e}\\ &=(n+1)\int_0^1\left[(1-t)^{n+1}+t^{n+1}\right]\,\mathrm{d}t\tag{1f}\\ &=\bbox[5px,border:2px solid #C0A000]{2\frac{n+1}{n+2}}\tag{1g} \end{align} $$ Explanation:
$\text{(1a)}$: $\binom{n}{k}=\frac{n!}{k!\,(n-k)!}$ and $n!=\Gamma(n+1)$
$\text{(1b)}$: $\Gamma(n+2)=(n+1)\Gamma(n+1)$
$\text{(1c)}$: definition of the Beta function in terms of the Gamma function
$\text{(1d)}$: definition of the Beta function in terms of an integral
$\text{(1e)}$: $\frac{1+x^{n+1}}{1+x}=1-x+x^2-\dots+x^n$ when $n$ is even
$\text{(1f)}$: simplify the integrand
$\text{(1g)}$: integrate

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A more elementary approach:

For $k\gt0$, partial fractions gives $$ \frac1{\binom{n}{k}}=\sum_{j=1}^k(-1)^{k-j}\binom{k}{j}\frac j{n-j+1}\tag{2} $$ Therefore, $$ \begin{align} \sum_{k=0}^n\frac{(-1)^k}{\binom{n}{k}} &=1+\sum_{k=1}^n\frac{(-1)^k}{\binom{n}{k}}\tag{3a}\\ &=1+\sum_{k=1}^n\sum_{j=1}^k(-1)^j\binom{k}{j}\frac j{n-j+1}\tag{3b}\\ &=1+\sum_{j=1}^n\sum_{k=j}^n(-1)^j\binom{k}{j}\frac j{n-j+1}\tag{3c}\\ &=1+\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac j{n-j+1}\tag{3d}\\ &=1+\sum_{j=0}^n(-1)^j\binom{n}{j}\frac{(n+1)\,j}{(j+1)(n-j+1)}\tag{3e}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n}{j}\left(\frac j{j+1}+\frac j{n-j+1}\right)\tag{3f}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n}{j}\frac n{j+1}\tag{3g}\\ &=1+\frac{n+1}{n+2}\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac n{n+1}\tag{3h}\\ &=1+\frac n{n+2}\tag{3i}\\ &=\bbox[5px,border:2px solid #C0A000]{2\frac{n+1}{n+2}}\tag{3j} \end{align} $$ Explanation:
$\text{(3a)}$: separate the $k=0$ term
$\text{(3b)}$: apply $(2)$
$\text{(3c)}$: change order of summation
$\text{(3d)}$: $\sum\limits_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}$
$\text{(3e)}$: $\binom{n+1}{j+1}=\binom{n}{j}\frac{n+1}{j+1}$
$\text{(3f)}$: partial fractions
$\text{(3g)}$: substitute $j\mapsto n-j$ in the right summand
$\text{(3h)}$: $\binom{n}{j}\frac1{j+1}=\binom{n+1}{j+1}\frac1{n+1}$
$\text{(3i)}$: $\sum\limits_{j=0}^n(-1)^j\binom{n+1}{j+1}=1$
$\text{(3j)}$: simplify

$$\underline{\text{Method 1}}$$

$$\text{S}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}$$

$$=\displaystyle\sum_{r=0}^{n}(-1)^r\dfrac{r!(n-r)!}{n!}$$

$$=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[(-1)^r\left\{r!(n-r+1)!+(n-r)!(r+1)!\right\}\right]$$

$$\color{blue}{=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[\left\{T_{r} - T_{r+1}\right\}\right]}\tag{*}$$

where $T_{r}= (-1)^r r!(n-r+1)!$

Clearly, $(*)$ is a telescoping series. Evaluating it, we have,

$$\text{S}=(-1)^nT_{n+1}+T_{0}$$

$$=(-1)^n\dfrac{(n+1)!}{n!(n+2)} + \dfrac{(n+1)!}{n!(n+2)}$$

Since $n$ is even,

$$\color{red}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}$$

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$$\underline{\text{Method 2}}$$

Consider the following Lemma.

Lemma : $$\color{brown}{\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}=0}$$

where $k$ is an odd positive integer.

Proof : $$\text{S}=\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}$$

$$=\displaystyle \sum_{r=0}^k\dfrac{(-1)^{k-r}}{\dbinom{k}{k-r}}$$

$$\displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r) = \sum_{r=0}^{n} f(n-r) \right)} $$

$$= -\displaystyle \sum_{r=0}^k\dfrac{(-1)^r}{\dbinom{k}{r}}$$

$$\implies \text{S}=-\text{S}$$

$$\implies \text{S}=0$$

This completes the proof of the Lemma.

Now, since $n$ is even, $n+1$ is odd.

Using the Lemma, we have,

$$\color{blue}{\displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}=0} \tag{1}$$

Let,

$$\color{red}{\text{J}=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}}}\tag{2}$$

Operating $(1)+(2)$,

$$\implies \text{J}+0=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n+1}\dfrac{(-1)^r}{\dbinom{n+1}{r}}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n+1}{r}} +\dfrac{(-1)}{\dbinom{n+1}{n+1}}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{n-r}}{\dbinom{n+1}{n-r}} -1 \quad \displaystyle\color{blue}{\left(\because \sum_{r=0}^{n} f(r)=\sum_{r=0}^{n} f(n-r)\right)}$$

$$=\displaystyle \sum_{r=0}^n\dfrac{(-1)^r}{\dbinom{n}{r}} + \displaystyle \sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n+1}{r+1}} -1$$

$$=\displaystyle \sum_{r=0}^n(-1)^r\left\{\dfrac{1}{\dbinom{n}{r}} + \dfrac{r+1}{n+1} \cdot \dfrac{1}{\dbinom{n}{r}}\right\} -1 \quad \displaystyle\color{red}{\left[\because \dbinom{n}{r}=\dfrac{n}{r}\dbinom{n-1}{r-1} \ \text{&} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]}$$

$$\implies \text{J} = \displaystyle\left(\dfrac{n+2}{n+1}\right) \sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}+ \dfrac{1}{n+1} \sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}} -1$$

$$\implies \text{J}+1 = \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{1}{n+1} \underbrace{\color{#66f}{\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}}}_{\huge{= \ \text{G} \ (\text{let})}} $$

$$\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{G}}{n+1} \tag{3} $$

Now,

$$\text{G}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r\cdot r}{\dbinom{n}{r}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{n-r}\cdot (n-r)}{\dbinom{n}{n-r}}$$

$$=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot (n-r)}{\dbinom{n}{r}}$$

$$=\displaystyle n\sum_{r=0}^{n}\dfrac{(-1)^{r}}{\dbinom{n}{r}} - \sum_{r=0}^{n}\dfrac{(-1)^{r}\cdot r}{\dbinom{n}{r}}$$

$$\implies \text{G}=n\cdot \text{J}-\text{G}$$

$$\implies\color{red}{\text{G}=\dfrac{\text{J}n}{2}} \tag{4}$$

From $(3)\ \text{&} \ (4)$,

$$\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}$$

$$\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square$$

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 $\int_{0}^{1}x(1-x)^n\,dx$.