Sum with binomial coefficients: $\sum_{k=1}^m  \frac{1}{k}{m \choose k} $ I got this sum, in some work related to another question:
$$S_m=\sum_{k=1}^m  \frac{1}{k}{m \choose k} $$
Are there any known results about this (bounds, asymptotics)? 

Consider a random task as follows.  First, one chooses a nonempty subset $X$ of $\{1,2,\ldots,m\}$, each with equal probability.  Then, one uniformly randomly selects an element $n$ of $X$.  The event of interest is when $n=\max(X)$.
Fix $k\in\{1,2,\ldots,m\}$.  The probability that $|X|=k$ is $\frac{1}{2^m-1}\,\binom{m}{k}$.  The probability that the maximum element of $X$, given $X$ with $|X|=k$, is chosen is $\frac{1}{k}$. Consequently, the probability that the desired event happens is given by $$\sum_{k=1}^m\,\left(\frac{1}{2^m-1}\,\binom{m}{k}\right)\,\left(\frac{1}{k}\right)=\frac{S_m}{2^m-1}\,.$$
Now, consider a fixed element $n\in\{1,2,\ldots,m\}$.  Then, there are $\binom{n-1}{j-1}$ possible subsets $X$ of $\{1,2,\ldots,m\}$ such that $n=\max(X)$ and $|X|=j$.  The probability of getting such an $X$ is $\frac{\binom{n-1}{j-1}}{2^m-1}$.  The probability that $n=\max(X)$, given $X$, is $\frac{1}{j}$.  That is, the probability that $n=\max(X)$ is $$\begin{align}
\sum_{j=1}^{n}\,\left(\frac{\binom{n-1}{j-1}}{2^m-1}\right)\,\left(\frac{1}{j}\right)
&=\frac{1}{2^m-1}\,\sum_{j=1}^n\,\frac1j\,\binom{n-1}{j-1}=\frac{1}{2^m-1}\,\left(\frac{1}{n}\,\sum_{j=1}^n\,\frac{n}{j}\,\binom{n-1}{j-1}\right)
\\
&=\frac{1}{2^m-1}\,\left(\frac{1}{n}\,\sum_{j=1}^n\,\binom{n}{j}\right)=\frac{1}{2^m-1}\left(\frac{2^n-1}{n}\right)\,.
\end{align}$$
Finally, it follows that $\frac{S_m}{2^m-1}=\sum_{n=1}^m\,\frac{1}{2^m-1}\left(\frac{2^n-1}{n}\right)$.  Hence, 
$$\sum_{k=1}^m\,\frac{1}{k}\,\binom{m}{k}=S_m=\sum_{n=1}^m\,\left(\frac{2^n-1}{n}\right)\,.$$
It can then be shown by induction on $m>3$ that $\frac{2^{m+1}}{m}<S_m<\frac{2^{m+1}}{m}\left(1+\frac{2}{m}\right)$.
