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?
Best Answer
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,
$$\sum_{n=4}^\infty\sum_{k=2}^{n-2} \binom{n}{k}^{-1} = \int_0^1 (1 + 3t(1-t)) dt = \frac32$$