Help in finding $\lim_{n\to\infty}\Bigl( \sum_{k=1}^{n} \frac{1}{({n \atop k})} \Bigr)^n$.

Question

I am not able to get a solution for this problem . Of finding the limit

$$\lim_{n\to\infty} \left( \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } \right)^n$$

I have tried using Mathematica and that numerically evaluates it to $7.3890560989 \cdots$

Which motivates me to think it is $e^2$ . Thanks for help

Answer 1

The largest summands are $\frac1{n\choose n}+\frac1{n\choose n-1}+\frac1{n\choose 1}=1+\frac2n$. The next smallest are $\frac1{n\choose n-2}=\frac1{n\choose 2}=\frac2{n(n-1)}=O(n^{-2})$, the others are $\le \frac1{n\choose 3}=\frac6{n(n-1)(n-2)}$. Since there are $n-4$ of these summands, their total is $\le \frac{6(n-4)}{n(n-1)(n-2)}=O(n^{-2})$. Hence the sum in parentheses is $1+\frac2n+O(n^{-2})$. As $\lim_{n\to\infty}(1+\frac xn)^n=e^x$, your guess was right, the limit is $e^2$.

Answer 2

$$\left(\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right )^n=e^{n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}}.$$ $$\binom{n}{1}=\binom{n}{n-1}=n,\binom{n}{2}=\binom{n}{n-2}=\frac{n(n-1)}{2},\binom{n}{2}\leq\binom{n}{k}, \ k=2,\cdots,n-2,$$ $$\sum_{k=1}^{n-1}\frac{1}{\binom{n}{k}}=\frac{2}{n}+\sum_{k=2}^{n-2}\frac{1}{\binom{n}{k}}<\frac{2}{n}+\frac{n-3}{\binom{n}{2}}\ \to0\ (n\to\infty).$$

$$n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}=n\log\left(1+\sum\limits_{k=1}^{n-1}\frac{1}{\binom{n}{k}}\right) \sim n\cdot\sum\limits_{k=1}^{n-1} \frac{1}{\binom{n}{k}}.$$ So $$\frac{n}{\binom{n}{2}}=\frac{n}{\binom{n}{n-2}}\to0,\ n\cdot\sum\limits_{k=3}^{n-3} \frac{1}{\binom{n}{k}}\leq\frac{n(n-5)}{\binom{n}{3}}\to 0\ (n\to\infty),$$ $$n\cdot\sum\limits_{k=1}^{n-1} \frac{1}{\binom{n}{k}}\to 2\ (n\to\infty).$$ $$\left(\sum_{k=1}^{n} \frac{1}{\binom{n}{k}}\right )^n=e^{n\log\sum\limits_{k=1}^{n} \frac{1}{\binom{n}{k}}}\to e^2\ (n\to\infty).$$

Answer 3

Using the following previous results we have that

• Calculate sums of inverses of binomial coefficients

$\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k} \implies \sum_{k=1}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}-1\tag{1}$

• Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$
• Is there any closed form expression for the following sum $\sum_{t=1}^T \frac{2^t}{t} = ?$

$$\sum_{k=1}^{n+1}\frac{2^k}{k} = \frac{2^{n+2}}{n+1}+\frac{2^{n+2}}{(n+1)^2}+O\left(\frac1n\right) \tag 2$$

therefore

$$ \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } = 1+\frac2{n+1}+O\left(\frac1{2^n}\right)$$

and therefore

$$\left( \sum_{k=1}^{n} \frac{1}{\binom{n}{k} } \right)^n = \left(1+\frac2{n+1}+O\left(\frac1{2^n}\right) \right)^n \to e^2$$