What is the asymptotic behavior of A103213 in OEIS? It's probably not at all hard—but at least right now it's not obvious to me—how to determine the asymptotic behavior of
$\sum_{k=1}^n \binom{n}{k} \frac{1}{k}$
(link to OEIS).
 A: Here's a heuristic argument using probability. 
Multiplying the OP's sum by $n/2^n$ gives 
$${1\over 2^n}\sum_{k=1}^n {n\choose k}{n\over k}=E\left({1\over \bar X_n}\right)$$
where $\bar X_n$ is the sample average of $n$ independent Bernoulli random variables with mean $1/2$ and we ignore the outcome $\bar X_n=0$. 
By the law of large numbers, $1/\bar X_n\to 2$ almost surely, making it plausible that  the left hand side of the equation above is approximately equal to 2.      
A: (see the related question)
Asymptotically, the sum behaves like the integral $$\sum_{k=1}^n \binom{n}{k} \frac{1}{k} = \int_1^n dk \, \frac{n!}{k! (n-k)! \,k}.$$ For large $n$, you can approximate the integral by expanding the integrand around its maximum (attained at $k=n/2$). We have (using Stirling) 
$$\log \frac{n!}{k! (n-k)!}  \sim n \log n - k \log k -(n-k) \log (n-k)$$
with the maximum at $k=n/2$. The expansion around $k=n/2$ reads
$$n \log n - k \log k -(n-k) \log (n-k) =  - \frac{2}{n}  (k- n/2)^2.$$
 Thereby, we can approximate
$$\sum_{k=1}^n \binom{n}{k} \frac{1}{k} \sim 
\frac{2}{n}\binom{n}{n/2} \int_1^n dk e^{ -2 (k- n/2)^2/n} \sim \frac{2}{n}
\frac{2^n}{\sqrt{\pi n/2}} \sqrt{\frac{\pi n}{2}}
=\frac{2^{n+1}}{n}
,$$
where we used the fact that $\binom{n}{n/2} \sim 2^n/\sqrt{\pi n/2}$. 
A: Here is an exact lower bound, which, as is readily seen, approximately equal to $2^{n+1}/n$.
By Jensen's inequality, since $1/x$ is convex, 
$$
\frac{{\sum\nolimits_{k = 1}^n {{n \choose k}} }}{{\sum\nolimits_{k = 1}^n {{n \choose k}} k}} \leq \frac{{\sum\nolimits_{k = 1}^n {{n \choose k}\frac{1}{k}} }}{{\sum\nolimits_{k = 1}^n {{n \choose k}} }}.
$$
From this it follows straightforwardly that
$$
\frac{{(2^n  - 1)^2 }}{{2^{n - 1} n}} \le \sum\limits_{k = 1}^n {{n \choose k}\frac{1}{k}} .
$$
EDIT: Hence,
$$
\sum\limits_{k = 1}^n {{n \choose k}\frac{1}{k}} \geq \frac{{2^{n + 1} }}{n} - \frac{4}{n} + \frac{1}{{2^{n - 1} n}}.
$$
A: Notice that $f(n) = \sum_{k=1}^n {n \choose k} \frac{1}{k} = \int_0^1 \frac{(t+1)^n - 1}{t}\, dt = \int_1^2 \frac{s^n - 1}{s-1} \, ds = \sum_{j=1}^{n} \frac{2^{j} - 1}{j}$.
I think the leading term should be $2^{n+1}/n$ 
