Evaluate $\sum_{n,k} \binom{n}{k}^{-1} $ 
Evaluate $$\sum_{n,k}  \frac{1}{\binom{n}{k}}, $$ where the summation ranges over all positive integers $n,k$ with $1<k<n-1$.

Thouhgts:
We are trying to evaluate $$\sum_{n=4}^{\infty} \sum_{k=2}^{n-2} \binom{n}{k}^{-1}$$
We may try to find a closed form of the inner summation which is of the form :
$$ \frac{1}{ {n \choose 2} } +  \frac{1}{{n \choose 3} }+ \dotsb + \frac{1}{{n \choose n-2} }.   $$
Notice that we may write $\frac{1}{ {n \choose 2} } = \frac{2! }{n(n-1)}$ and if keep doing the same for the other terms we obtain the following:
$$ \frac{ (n-2)! + (n-3)! (n-(n-2)) + (n-4)!(n-(n-2))(n-(n-3)) + \dotsb + 2! (n-3)! }{n!}, $$
which equals
$$ \frac{ (n-2)! + 2!(n-3)! + 3! (n-4)! + \dotsb + (n-3)! 2! }{n!} $$
and so this equals:
$$ \frac{1}{n(n-1)} + \frac{2}{n(n-1)(n-2)} + \dfrac{6}{n(n-1)(n-2)} + \dotsb + \dfrac{2}{n(n-1)(n-2) }. $$
But half of this terms are identical. Therefore, we are trying to sum up series of the form
$$\sum_{n \geq k} \frac{1}{(n-1)(n-2)(n-3)\dotso(n-k)} ,$$
which can be done by a telescoping trick, but it seems very formidable. Am I approaching this problem the right way? Any hints/suggestions?
 A: Partial Fractions and Telescoping Sums
$$
\begin{align}
\sum_{n=4}^\infty\sum_{k=2}^{n-2}\frac1{\binom{n}{k}}
&=\sum_{k=2}^\infty\sum_{n=k+2}^\infty\frac1{\binom{n}{k}}\tag1\\
&=\sum_{k=2}^\infty\frac{k}{k-1}\sum_{n=k+2}^\infty\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}\right)\tag2\\
&=\sum_{k=2}^\infty\frac{k}{k-1}\frac1{\binom{k+1}{k-1}}\tag3\\
&=\sum_{k=2}^\infty\frac2{(k-1)(k+1)}\tag4\\
&=\sum_{k=2}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)\tag5\\[3pt]
&=1+\frac12\tag6\\[9pt]
&=\frac32\tag7
\end{align}
$$
Explanation:
$(1)$: change order of summation
$(2)$: $\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}=\frac{\frac nk}{\binom{n}{k}}-\frac{\frac{n-k+1}k}{\binom{n}{k}}=\frac{\frac{k-1}k}{\binom{n}{k}}$
$(3)$: telescoping sum
$(4)$: $\binom{k+1}{k-1}=\binom{k+1}{2}=\frac{(k+1)k}2$
$(5)$: partial fractions
$(6)$: telescoping sum
$(7)$: simplify
A: Let $\ell = n -k$, we have
$$\begin{align} \sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1}
&= \sum_{k=2}^\infty\sum_{n=k+2}^\infty \binom{n}{k}^{-1}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \binom{k+\ell}{k}^{-1}
= \sum_{k=2}^\infty\sum_{\ell=2}^\infty \frac{k!\ell!}{(k+\ell)!}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)\frac{\Gamma(k+1)\Gamma(\ell+1)}{\Gamma(k+\ell+2)}\\
&= \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)\int_0^1 t^k (1-t)^\ell dt\\
&= \int_0^1 \sum_{k=2}^\infty\sum_{\ell=2}^\infty \left[ (k+\ell+1) t^k (1-t)^\ell\right] dt
\end{align}
$$
Notice when $s$ and $t$ are independent, we have
$$\begin{align} \sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1)t^k s^\ell
= & \sum_{k=2}^\infty\sum_{\ell=2}^\infty \left(t\frac{\partial}{\partial t} + s\frac{\partial}{\partial s} + 1 \right)t^k s^\ell
\\
= & \left(t\frac{\partial}{\partial t} + s\frac{\partial}{\partial s} + 1 \right)
\frac{t^2s^2}{(1-t)(1-s)}\\
= & \frac{s^2t^2(5-4(s+t)+3st)}{(1-s)^2(1-t)^2}
\end{align}$$
Substitute $s$ by $1-t$, we obtain
$$\sum_{k=2}^\infty\sum_{\ell=2}^\infty (k+\ell+1) t^k (1-t)^\ell = 1 + 3t(1-t)$$
As a result,
$$\sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1} = \int_0^1 (1 + 3t(1-t)) dt = \frac32$$
A: Generally similar sums can be evaluated using the Beta function:
$$
B(x+1,y+1)=\int_0^1 t^{x}(1-t)^{y}dt=\frac{\Gamma(x+1)\Gamma(y+1)}{\Gamma(x+y+2)}=
\frac{x!y!}{(x+y+1)!}=\frac{1}{x+y+1}\binom{x+y}x^{-1}.
$$ 
Applying this in your case ($x=n-k,y=k$) one has:
$$
\binom{n}{k}^{-1}=(n+1)\int_0^1 t^{n-k}(1-t)^{k}dt
$$
or
$$\begin{align}
I_n=\frac1{n+1}\sum_{k=2}^{n-2}\binom{n}{k}^{-1}
&=\sum_{k=2}^{n-2}\int_0^1 t^{n-k}(1-t)^kdt\\
&=\int_0^1\left[ t^n  \sum_{k=2}^{n-2}\left(\frac{1-t}t\right)^k\right] dt\\
&=\int_0^1 t^n\left(\frac{1-t}t\right)^2
\frac{1-\left(\frac{1-t}t\right)^{n-3}}{1-\frac{1-t}t}dt\\
\end{align}$$
and, finally,
$$\begin{align}
\sum_{n=4}^\infty (n+1)I_n&
=\int_0^1\left[\frac{(1-t)^2}{2t-1}\sum_{n=4}^\infty (n+1)\left(t^{n-1}-t^2(1-t)^{n-3}\right)\right]dt\\
&=\int_0^1(1+3t-3t^2)dt=\frac32.
\end{align}$$
