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)? 

You know that $\displaystyle (x + 1)^m = \sum_{k=0}^m {m \choose k} x^k$. So 
$$\int_0^1 \frac{(x + 1)^m - 1}{x} \, dx = \sum_{k=1}^m {m \choose k} \frac{1}{k}.$$
Letting $y = x + 1$ this is just 
$$\int_1^2 \frac{y^m - 1}{y - 1} \, dy = \sum_{k=1}^m \frac{2^k - 1}{k}.$$
The contribution of the $-1$ terms is $-H_m \sim - \log m$, so let's concentrate on estimating
$$T_m = \sum_{k=1}^m \frac{2^k}{k}.$$
There is an obvious lower bound
$$T_m \ge \sum_{k=1}^m \frac{2^k}{m} = \frac{2^{m+1} - 2}{m}.$$
To get an upper bound, we'll split the sum as
$$\sum_{k=1}^{m-r} \frac{2^k}{k} + \sum_{k=m-r+1}^m \frac{2^k}{k}$$
for some $r$ (depending on $m$, although we do not specify the dependence for now). (Thanks to leonbloy for improvements in the part of the argument that follows!) Noting that $f(k) = \frac{2^k}{k}$ is an increasing function of $k$, the first part is bounded above by $(m-r) \frac{2^{m-r}}{m-r} = 2^{m-r}$ while the second is bounded above by 
$$\sum_{k=m-r+1}^m \frac{2^k}{m-r+1} \le \frac{1}{m-r+1} \sum_{k=0}^m 2^k < \frac{2^{m+1}}{m-r+1}.$$
The first part gets smaller as $r$ increases while the second gets larger; to minimize their sum it is generally a good idea to make the two parts about as equal as possible. Thus we want $2^{m-r} \approx \frac{2^{m+1}}{m-r+1}$, or
$$(m-r+1) \approx 2^{r+1}.$$
This gives $r \approx \log_2 m$. Setting $r = (1 + \epsilon) \log_2 m$ for some $\epsilon > 0$ makes the first part negligible relative to the second without really increasing the second and together with the lower bound gives an asymptotic
$$T_m \sim \frac{2^{m+1}}{m}.$$
Edited by leonbloy: With respect to the upper bound - we have that, for any $r = 1..m$:
$$T_m \le 2^{m-r} + \frac{2^{m+1}}{m-r+1} = 2^m \left(2^{-r} + \frac{2}{m-r+1}\right)$$
The expression in parentheses, thought as a function of $r$ (continuous), has a global minimum at $ 2^{r+1} = (m+1-r)^2 \log(2)$, which for large $m$ would give $r\approx 2 \log_2(m)$. Inspired by this, we can choose $r=\lceil 2 \log_2(m) \rceil$,
and hence we can bound the two terms:
$$r \ge 2 \log_2(m) \Rightarrow    2^{-r}  \le \frac{1}{m^2}$$
$$r \le 2 \log_2(m) +1 \Rightarrow \frac{2}{m-r+1} \le  \frac{2}{m-2 \log_2(m) }$$
So, finally we have the bounds:
$$  \frac{2}{m}(1+ 2^{-m}) \le \frac{T_m}{2^{m}} \le \frac{2}{m-2 \log_2(m) } +\frac{1}{m^2} $$
which, agrees with the above asymptotic.
