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?

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
