# 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!

• Does the series telescope? Sep 17, 2020 at 2:13
• Perhaps you've seen something like $\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1$, using partial fractions on the summands and then telescoping (often used as an example in calculus courses). Try $k=3,4$ and see what happens.
– yoyo
Sep 17, 2020 at 2:21

\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

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?

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, $$S=\frac{k}{k-1}, k\ne 1$$