Q:

Evaluate $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$

I'm interested in finding a nice closed form expression for the sum $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$. I've tried using the Binomial Theorem to get 
\begin{align*}
\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k & =\int_0^1\frac{(1+x)^n-1}{x} \, dx\\
&=\int_1^2 (1+u+\cdots+u^{n-1}) \, du
\end{align*}
using the substitution $u=1+x$ but I can't quite to simplify this integral either.  I have also not been able to come up with a combinatorial approach, which may not exist since the summation and its terms are in general not integers.  Any help in evaluating this sum would be appreciated, thanks! 

A:

In the question, the problem is in "nice closed form expression" since
$$\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k=n x \, _3F_2(1,1,1-n;2,2;-x)$$ where appears an hypergoemetric function. 
So, let us forget the $x$ and compute for a few values of $n$
$$S_n=\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$$
$$\left\{1,\frac{5}{2},\frac{29}{6},\frac{103}{12},\frac{887}{60},\frac{1517}{60},\frac{18239}{420}\right\}$$ which are
$$\left\{1,\frac{5}{2},\frac{29}{6},\frac{206}{24},\frac{1774}{120},\frac{18204}{720},\frac{218868}{5040}\right\}$$ The denominators are clearly $n!$ and the numerators corresponds to sequence A103213 in OEIS. 
As you will see in the link is that, for large $n$
$$S_n\approx \frac{2^{n+1}} n$$ It is also given that
$$S_n=-H_n-\Re(B_2(n+1,0))$$ where appear the harmonic number and the real part of the  incomplete beta function. 
Update
Concerning the asymptotic behavior, it seems that it could be slightly improved using
$$S_n\approx 2^{n+1} n^{\frac{1}{4 n}-1}$$

