Limiting the Inverse Binomial Sum

Given: $\text{S}(n) = \displaystyle \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$

$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$

$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}$ $\left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$

Also,

$\text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}}$ $\left[ \text{since} \ \displaystyle \sum_{r=a}^b f(r) = \displaystyle \sum_{r=a}^b f(a+b-r) \ \text{and} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]$

Thus, we have,

$\begin{cases} \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}\\ \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\binom{n-1}{r-1}} \end{cases}$

Adding the above two expressions, we get,

$2\text{S}(n) = 2 + \displaystyle \sum_{r=1}^n \left( \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} + \dfrac{n-r+1}{n} \dfrac{1}{\binom{n-1}{r-1}} \right)$

$= 2 + \dfrac{n+1}{n} \displaystyle \sum_{r=1}^n \dfrac{1}{\binom{n-1}{r-1}}$

$= 2 + \dfrac{n+1}{n} \times \text{S}(n-1)$

Since $n \to \infty$ , we have $\text{S}(n) = \text{S}(n-1) = \text{S}$ (say)

$\implies 2\text{S} = \left(\dfrac{n+1}{n}\right) \times \text{S} +2$

$\implies \text{S} = \dfrac{2n}{n-1} = \boxed{2}$ [since $n \to \infty$ ]

NOTE: For the existance of the limit, we can easily prove that $\text{S}(n+1) < \text{S}(n)$ for $n \ge 4$ , hence the limit exists.

This problem can easily be tackled using Squeeze theorem :). First of all,note that, for all numbers, $\binom{n}{2} \le \binom{n}{r} \forall r \in [2,n-2]$ Now, $\sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = 1+\frac{1}{n}+ \sum_{r=2}^{n-2} \frac{1}{\binom{n}{r}} + \frac{1}{n}+1 \ge 2$ Also, $\sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = 1+\frac{1}{n}+ \sum_{r=2}^{n-2} \frac{1}{\binom{n}{r}} + \frac{1}{n}+1 \le 2+ \frac{2}{n} + \sum_{r=2}^{n-2} \frac{1}{\binom{n}{2}}$ And, $2+ \frac{2}{n} + \sum_{r=2}^{n-2} \frac{1}{\binom{n}{2}} = 2+ \frac{2}{n} + \sum_{r=2}^{n-2} \frac{2}{n(n-1)} = 2+ \frac{2n-6+2n-2}{n(n-1)} = 2+\frac{4n-8}{n(n-1)}$

Hence, $2 \le \sum_{r=0}^{n} \frac{1}{\binom{n}{r}} \le 2+\frac{4n-8}{n(n-1)}$

Taking limit, and using statement of squeeze theorem, $\lim_{n \rightarrow \infty} S_n = 2$