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!
Best Answer
$$ \begin{align} \sum_{n=k}^\infty\frac1{\binom{n}{k}} &=\sum_{n=k}^\infty\frac{k!}{n(n-1)\cdots(n-k+1)}\tag1\\ &=\sum_{n=k}^\infty\frac{k!}{k-1}\left({\scriptsize\frac{n}{n(n-1)\cdots(n-k+1)}-\frac{n-k+1}{n(n-1)\cdots(n-k+1)}}\right)\tag2\\ &=\sum_{n=k}^\infty\frac{k!}{k-1}\left({\scriptsize\frac1{(n-1)(n-2)\cdots(n-k+1)}-\frac1{n(n-1)\cdots(n-k+2)}}\right)\tag3\\ &=\sum_{n=k}^\infty\frac{k}{k-1}\left({\scriptsize\frac{(k-1)!}{(n-1)(n-2)\cdots(n-k+1)}-\frac{(k-1)!}{n(n-1)\cdots(n-k+2)}}\right)\tag4\\ &=\sum_{n=k}^\infty\frac{k}{k-1}\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}\right)\tag5\\ &=\frac{k}{k-1}\tag6 \end{align} $$ Explanation:
$(1)$: expand the binomial coefficient
$(2)$: $\frac{n-(n-k+1)}{k-1}=1$
$(3)$: cancel fractions
$(4)$: distribute $(k-1)!$
$(5)$: collect binomial coefficients
$(6)$: the sum telescopes