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

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 $$S_n\approx 2^{n+1} n^{\frac{1}{4 n}-1}$$

Here's a way to get the same result, and another form as well, without integrals: $$ \begin{align} f(n) &=\sum_{k=1}^n\frac1k\binom{n}{k}\\ &=\sum_{k=1}^n\frac1k\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]\\ &=f(n-1)+\sum_{k=1}^n\frac1k\binom{n-1}{k-1}\\ &=f(n-1)+\frac1n\sum_{k=1}^n\binom{n}{k}\\ &=f(n-1)+\frac{2^n-1}n\\ &=\bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^n\frac{2^k-1}k}\\ &=\sum_{k=1}^n\frac1k\sum_{j=0}^{k-1}2^j\\ &=\sum_{j=0}^{n-1}2^j\sum_{k=j+1}^n\frac1k\\ &=\bbox[5px,border:2px solid #C0A000]{\sum_{j=0}^{n-1}2^j(H_n-H_j)} \end{align} $$ However, I don't see a simple closed form.