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?

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,
$$\begin{align}
\sum_{n=4}^\infty (n+1)I_n&
=\int_0^1\left[\frac{(1-t)^2}{2t-1}\sum_{n=4}^\infty (n+1)\left(t^{n-1}-t^2(1-t)^{n-3}\right)\right]dt\\
&=\int_0^1(1+3t-3t^2)dt=\frac32.
\end{align}$$
