Series of inverses of binomial coefficients Can you think of a simple way of proving that
$$
\sum_{n=k+1}^\infty \frac{1}{n \choose k}
$$ 
is rational for any $k \geq 2$?
Here's the background. Consider a series:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)}
$$
Elementary algebra gives us that:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = \lim_{k\to\infty} (1 - \frac{1}{k+1}) = 1
$$
so the sum turns out to be rational.
Next, consider 
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)}
$$
With the same method, but much more effort we can show that:
$$
\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)(n+3)(n+4)} = \sum_{n=1}^\infty \left(\frac{1}{24 n}-\frac{1}{6(n+1)}+\frac{1}{4(n+2)}-\frac{1}{6(n+3)}+\frac{1}{24(n+4)}\right)
$$
and again we'll see that stuff cancels out, and the sum is again rational.
So the obvious conjecture is that this method will work for arbitrary (but $\geq 2$) fixed number of factors in denominator, and the sum will always be rational. Indeed that's the case. I provide a solution as an answer, but I'm not fully satisfied with it (it seems for me to be too brute force), so I'm looking for alternative solutions.


We show the result stated by @IgorRivin is valid.
\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{\binom{n}{k}}=\frac{1}{k-1}\qquad\qquad k\geq 2\tag{1}
\end{align*}

We start proving a related identity from which we can easily derive (1).

The following is valid for $N\geq 1,k\geq 1$
\begin{align*}
\sum_{n=1}^{N}\frac{1}{\binom{n+k+1}{k+1}}=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)}{\binom{N+k+1}{k}}\tag{2}
\end{align*}

We obtain
\begin{align*}
\sum_{n=1}^{N}\frac{1}{\binom{n+k+1}{k+1}}&=\sum_{n=1}^{N}\frac{n!(k+1)!}{(n+k+1)!}\\
&=(k+1)!\sum_{n=1}^N\frac{1}{(n+1)\cdots(n+k+1)}\\
&=\frac{(k+1)!}{k}\sum_{n=1}^N\frac{(n+k+1)-(k+1)}{(n+1)\cdots(n+k+1)}\\
&=\frac{(k+1)!}{k}\sum_{n=1}^N\left(\frac{1}{(n+1)\cdots(n+k)}-\frac{1}{(n+2)\cdots(n+k+1)}\right)\tag{3}\\
&=\frac{(k+1)!}{k}\left(\frac{1}{(k+1)!}-\frac{1}{(N+2)\cdots(N+k+1)}\right)\\
&=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)k!}{(N+2)\cdots(N+k+1)}\tag{4}\\
&=\frac{1}{k}-\frac{\left(1+\frac{1}{k}\right)}{\binom{N+k+1}{k}}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
In (3) we get a telescoping sum leaving only the first and last term.

We will now apply identity (2) to answer OPs question. We use the representation from (4) with $k-1$ instead of $k$ and the claim follows for $k\geq 2$:
\begin{align*}
\sum_{n=k+1}^{\infty}\frac{1}{\binom{n}{k}}&=\sum_{n=1}^{\infty}\frac{1}{\binom{n+k}{k}}
=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}\frac{1}{\binom{n+k}{n}}\\
&=\lim_{N\rightarrow\infty}\left(\frac{1}{k-1}-\frac{\left(1+\frac{1}{k-1}\right)(k-1)!}{(N+2)\cdots(N+k)}\right)\\
&=\frac{1}{k-1}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}

Note: These formulae can be found in H.W.Goulds Combinatorial Identities Vol. 3, (3.26 - 3.29).
