Evaluating $\sum_{n=k}^{\infty} \frac{1}{ \binom{n}{k}}$ 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!
 A: $$
\begin{align}
\sum_{n=k}^\infty\frac1{\binom{n}{k}}
&=\sum_{n=k}^\infty\frac{k!}{n(n-1)\cdots(n-k+1)}\tag1\\
&=\sum_{n=k}^\infty\frac{k!}{k-1}\left({\scriptsize\frac{n}{n(n-1)\cdots(n-k+1)}-\frac{n-k+1}{n(n-1)\cdots(n-k+1)}}\right)\tag2\\
&=\sum_{n=k}^\infty\frac{k!}{k-1}\left({\scriptsize\frac1{(n-1)(n-2)\cdots(n-k+1)}-\frac1{n(n-1)\cdots(n-k+2)}}\right)\tag3\\
&=\sum_{n=k}^\infty\frac{k}{k-1}\left({\scriptsize\frac{(k-1)!}{(n-1)(n-2)\cdots(n-k+1)}-\frac{(k-1)!}{n(n-1)\cdots(n-k+2)}}\right)\tag4\\
&=\sum_{n=k}^\infty\frac{k}{k-1}\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom{n}{k-1}}\right)\tag5\\
&=\frac{k}{k-1}\tag6
\end{align}
$$
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
A: So the sum you are concerned with is $$\sum_{n=k}^{\infty} \frac{1}{(n-k+1) \cdot (n-k+2)\dots \cdot n}$$ which has the product of $k$ consecutive integers in the denominator of each summand which you can express as the difference of two fractions each with the product of $k-1$ consecutive terms in the denominator. So, for $n=N$, the summand is $$\dfrac{1}{(N-k+1)\cdots(N-1) N}\\ = \dfrac1{k-1}\left(\dfrac{1}{(N-k+1)(N-k+2)\cdots (N-1)}-\dfrac{1}{(N-k+2)\cdots(N-1)N}\right)$$

To get a feel of this fix a $k>2$, say $k=4$, and explicitly write down a few of the summands at the beginning, you should have $$\dfrac1{1.2.3.4}+\dfrac1{2.3.4.5}+\dfrac1{3.4.5.6}+\cdots$$ which using the decomposition mentioned above, becomes $$\frac13\left(\dfrac1{1.2.3}-\dfrac1{2.3.4}\right)+\frac13\left(\dfrac1{2.3.4}-\dfrac1{3.4.5}\right)+\frac13\left(\dfrac1{3.4.5}-\dfrac1{4.5.6}\right)+\cdots$$
do you see what happens if you take $\dfrac1{3}=\dfrac1{k-1}$ common out of all of them?
A: 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, $$S=\frac{k}{k-1}, k\ne 1$$
