Asymptotic rate for $\sum\binom{n}k^{-1}$

This sum can be written in the form, see 2-adic Logarithm and Resistance of n-dimensional Cube $$S_{n+1}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}.$$ The last term in the sum gives the estimate $S_n>1.$ You can get sharper estimates taking more terms.

The following asymptotic expansion is proved in https://arxiv.org/abs/0904.1757 (The Hypercube of Resistors, Asymptotic Expansions, and Preferential Arrangements, by Nicholas Pippenger. Published in Mathematics Magazine 83(N5) (2010), 331-346): $$\sum_{k=0}^n\frac{1}{\binom{n}{k}}=2\left(1+\frac{1}{n}+\frac{2}{n(n-1)}+\ldots \frac{k!}{n(n-1)\cdots(n-k+1)}+O(\frac{1}{n^{k+1}})\right).$$

P.S. This result confirms the answer given in the Gerhard Paseman's comment.