TITLE: Evaluating $\sum_{n=k}^{\infty} \frac{1}{ \binom{n}{k}}$
QUESTION [9 upvotes]: I've looked into WolframAlpha and deduced from some examples that:
$$ \sum_{n=k}^{\infty} \frac{1}{ \binom{n}{k}}  = \frac{k}{k-1} ~~~~ \text{where} ~~~ k \in \mathbb{N} \setminus \{1\}$$
But why is that? The only thing I could pull of is this:
$$ \sum_{n=k}^{\infty} \frac{1}{ \binom{n}{k}} = \sum_{n=k}^{\infty} \frac{k! (n-k)!}{n!} = k! \sum_{n=k}^{\infty} \frac{ (n-k)!}{n!} = k! \sum_{n=k}^{\infty} \frac{1}{(n-k+1) \cdot (n-k+2)\dots \cdot n}$$
Which then got me into a dead-end (for my knowledge) ... I am curious as why is that and but this actually mean "combinatorically" / "statistically" and how to actually evaluate this.
Thanks!

REPLY [4 votes]: Use $${n \choose k}^{-1}=(n+1) \int_{0}^{1} x^k (1-x)^{n-k} dx$$
So, $$S=\sum_{n=k}^{\infty} {n \choose k}^{-1}=\sum_{n=k}^{\infty}(n+1)\int_{0}^{1} x^k(1-x)^{n-k} dx=\int_{0}^{1}\sum_{n=k}^{\infty} (n+1) x^k (1-x)^{n-k} dx$$
Let $n-k=p$, then
$$S=\int_{0}^{1}x^k dx\sum_{p=0}^{\infty}  (p+k+1) (1-x)^p$$
Using $$\sum_{j=0}^{\infty} j~z^{j}=\frac{z}{(1-z)^2}, \sum_{j=0}^{\infty} z^j=\frac{1}{1-z}$$
We get $$S=\int_{0}^{1}\left( \frac{1-x}{x^2}+\frac{1+k}{x}\right) x^k dx=\int(x^{k-2}+k x^{k-1}) dx.$$
Finally, $$S=\frac{k}{k-1}, k\ne 1$$