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

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