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.
 A: 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 $\int_{0}^{1}x(1-x)^n\,dx$.
A: 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
$$\bbox[5px,border:2px solid #00A000]{
2 \frac{n+1}{n+2}.}$$
A: 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

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
A: $$\underline{\text{Method 1}}$$
$$\text{S}=\displaystyle\sum_{r=0}^{n}\dfrac{(-1)^r}{\dbinom{n}{r}}$$
$$=\displaystyle\sum_{r=0}^{n}(-1)^r\dfrac{r!(n-r)!}{n!}$$
$$=\displaystyle\sum_{r=0}^{n}(-1)^r \dfrac{r!(n-r)!\color{blue}{(n-r+1+r+1)}}{n!}\color{red}{\times\dfrac{1}{(n+2)}}$$
$$=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[(-1)^r\left\{r!(n-r+1)!+(n-r)!(r+1)!\right\}\right]$$
$$\color{blue}{=\displaystyle\sum_{r=0}^{n}\dfrac{1}{n!(n+2)}\left[\left\{T_{r} - T_{r+1}\right\}\right]}\tag{*}$$
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,
$$\color{red}{\therefore\text{S}=\boxed{2\dfrac{n+1}{n+2}}}$$

$$\underline{\text{Method 2}}$$
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 $(3)\ \text{&} \ (4)$,
$$\implies \text{J}+1= \displaystyle\left(\dfrac{n+2}{n+1}\right) \text{J}+ \dfrac{\text{J}n}{2(n+1)}$$
$$\color{green}{\therefore\text{J}=\boxed{2\dfrac{n+1}{n+2}}} \quad \square$$
A: HINT:
In Brychkov, Marichev, Prudnikov Vol 1, 4.2.4 #2 there appears the more general sum:
$$\sum_{k=0}^n (-1)^k\binom{a}{k}^{-1}$$
We use the following formula
$$\binom{a}{k}^{-1} = \frac{a+1}{a+2}\left(\binom{a+1}{k}^{-1} + \binom{a+1}{k+1}^{-1}\right)$$
Now the alternating sum of terms from the LHS becomes a telescoping sum.
$\bf{Added:}$ The above quoted volume contains many other interesting results.
