I'm trying to find a closed solution to the following binomial sum, without success. I would appreciate any assistance towards a solution.
$$
\sum\limits_{k=0}^{n-1}\dfrac{1}{\binom{n}{k}(n-k)}
$$
Perhaps the following alternative form presents the problem in a better light:
$$
\dfrac{1}{n!}\sum\limits_{k=0}^{n-1}k![(n-1)-k]!
$$
UPDATE:
Since Peter Košinár's and Phira's kind and informative responses, I have found some other equivalent expressions:
$$
\sum\limits_{k=0}^{n-1}\dfrac{1}{2^k (n-k)}
$$
and
$$
\dfrac{1}{2^n}\sum\limits_{k=1}^{n}\dfrac{2^k}{k}
$$
The first can be proved from the identity stated here (Wayback Machine).
The second I have seen merely stated, but cannot prove from my original expression.
Best Answer
Equation $(8)$ from this answer says $$ \sum_{m=0}^nm!(n-m)!=\frac{(n+1)!}{2^n}\sum_{k=0}^n\frac{2^k}{k+1}\tag{1} $$ Applying this to your sum gives $$ \begin{align} \sum_{k=0}^{n-1}\frac1{\binom{n}{k}(n-k)} &=\frac1{n!}\sum_{k=0}^{n-1}k!(n-k-1)!\\ &=\frac1{2^{n-1}}\sum_{k=0}^{n-1}\frac{2^k}{k+1}\\ &\sim\frac2n+\frac2{n(n-1)}+\frac4{n(n-1)(n-2)}\\ &+\frac{12}{n(n-1)(n-2)(n-3)}\\ &+\frac{48}{n(n-1)(n-2)(n-3)(n-4)}+O\left(\frac1{n^6}\right)\tag{2} \end{align} $$ where we used $(6)$ below to get the asymptotic expansion.
Asymptotic Expansion
Note that $$ \begin{align} &\frac{(n-j+1)!}{n!}\sum_{k=0}^{n-j}\frac{(k+j)!}{k!}\frac{2^{-k-j}}{n-j-k+1}\\ -&\frac{(n-j+1)!}{n!}\sum_{k=0}^{n-j}\frac{(k+j)!}{k!}\frac{2^{-k-j}}{n-j+1}\\ =&\frac{(n-j+1)!}{n!}\sum_{k=0}^{n-m}\frac{(k+j)!}{k!}\frac{k\,2^{-k-j}}{(n-j-k+1)(n-j+1)}\\ =&\frac{(n-(j+1)+1)!}{n!}\sum_{k=0}^{n-(j+1)}\frac{(k+(j+1))!}{k!}\frac{2^{-k-(j+1)}}{n-(j+1)-k+1}\tag{3} \end{align} $$ Thus, $$ \begin{align} (n+1)\sum_{k=0}^n\frac{2^{-k}}{n-k+1} &=\sum_{j=0}^{m-1}\frac{(n-j)!}{n!}\sum_{k=0}^{n-j}\frac{(k+j)!}{k!}2^{-k-j}\\ &+\frac{(n-m+1)!}{n!}\sum_{k=0}^{n-m}\frac{(k+m)!}{k!}\frac{2^{-k-m}}{n-m-k+1}\\ &=\sum_{j=0}^{m-1}\frac{2}{\binom{n}{j}}+O\left(\frac1{n^m}\right)\tag{4} \end{align} $$ since $$ \begin{align} \sum_{k=0}^\infty\frac{(k+j)!}{k!}2^{-k-j} &=j!\sum_{k=0}^\infty\binom{k+j}{k}2^{-k-j}\\ &=j!2^{-j}\sum_{k=0}^\infty(-1)^k\binom{-j-1}{k}2^{-k}\\ &=j!2^{-j}\left(1-\frac12\right)^{-j-1}\\[12pt] &=2j!\tag{5} \end{align} $$ Combining $(1)$ and $(4)$ yields $$ \sum_{m=0}^n\frac1{\binom{n}{m}}=\sum_{j=0}^{m-1}\frac{2}{\binom{n}{j}}+O\left(\frac1{n^m}\right)\tag{6} $$ Equation $(6)$ may seem like a big circle, but it actually tells us that we can ignore the sum of the middle terms as being of the order of the largest term omitted.