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.
 A: Consider the generating function of the terms of our series (after simplifying renumbering):
$$
F_k(x) = \sum_{n=k+1}^\infty \frac{x^n}{n(n-1)\cdots(n-k)}
$$
By explicit calculation, it's easy to see that $k-th$ derivative of $F_k$ is equal to $-\log(1-x)$, so indeed we can calculate $F_k$ by taking antiderivative of $-\log(1-x)$ $k$ times. Surprisingly, it actually leads us somewhere in our case: Inductive argument using integration by parts show that the $k$-th antiderivative of $-\log(1-x)$ is $C (1-x)^k \log(1-x) + W(x)$, where $W$ is some polynomial of degree $k$ and $C$ is some rational constant (the exact value of C and the polynomial $W$ are not really hard to obtain, but they're not important here). 
So, by general argument involving power series, we see that for $n > k$:
$$
\frac{1}{n(n-1)\cdots(n-k)} = \frac{F^{(n)}_k(0)}{n!}
$$
But then 
$$
\frac{F^{(n)}_k(x)}{n!} = C \frac{((1-x)^k log(1-x))^{(n)}}{n!} = C \sum_{i=0}^k (-1)^i {k \choose i} \frac{(x^i \log(1-x))^{(n)}}{n!}
$$
The polynomial $W$ disappears because $n$ is larger than degree of $W$. Now, $\frac{(x^i \log(1-x))^{(n)}}{n!}$ is precisely the $n$-th coefficient in the power series expansion of $x^i \log(1-x) = x^i(-x - \frac{1}{2} x^2 - \ldots - \frac{1}{n} x^n + \ldots) = -x^{i+1} - \frac{1}{2} x^{i+2} - \ldots - \frac{1}{n} x^{i+n} + \ldots$, so it's exactly $\frac{1}{n-i}$.
In the end, we obtain:
$$
\sum_{n=k+1}^\infty \frac{1}{n(n-1)\cdots(n-k)} = \sum_{n=k+1}^\infty \frac{F^{(n)}_k(0)}{n!}  = \sum_{n=k+1}^\infty  C \sum_{i=0}^k (-1)^i {k \choose i} \frac{1}{n-i}
$$
Using the fact that $\sum_{i=0}^k (-1)^i {k \choose i} = 0$ and appropriately reordering terms, we again see that most terms cancel out, and just like we saw for $k = 1$ and $k = 4$, sum is rational (remember that $C$ was rational). Bonus point is that since $C$ is easily computable (left as an exercise for reader), we can actually calculate the sum.
A: Mathematica gives the sum in closed form:
$$
\frac{\Gamma (k-1) \Gamma (k+2)}{(k+1) \Gamma (k) \Gamma (k+1)}.
$$
If you ask it to FullSimplify[] it, it comes up with 
$$\frac{1}{k-1}.$$
Here is the relevant command:
Assuming[Element[k, Integers], 
 FullSimplify[Sum[1/Binomial[n, k], {n, k + 1, Infinity}]]]
A: 
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).
