Asymptotic behavior of $\sum\limits_{k=1}^n \frac{2^k}{k}$ I'm looking for an asymptotic equivalent of
$$\sum_{0 < k \le n} \frac{2^k}{k}$$
as $n \to \infty$. A plausible candidate seems to be $\frac{2^{n+1}}{n+1}$ (WolframAlpha plot, and the intuitive similarity with $\sum_{k \le n} 2^k = 2^{n+1}$ is also appealing), but my usual tricks seem powerless here.
I've tried:


*

*Interpreting $x^k/k$ as the primitive of $x^{k-1}$ and setting $x=2$

*Replacing $2^k/k$ with $\int_{x=0}^2 x^{k-1}$

*Reordering the sum's terms to expose log-resembling sub-sums like $\sum_k 1/k$

*Finding lower and upper bounds asymptotically equivalent to $\frac{2^{n+1}}{n+1}$ -- the lower bound is easy ($\sum\limits_{k \le n}  \frac{2^k}{k} \ge \sum\limits_{k \le n} \frac{2^k}{n+1} \ge \frac{\sum\limits_{k \le n} 2^k}{n+1} \ge \frac{2^{n+1}}{n+1}$), but the upper bound seems trickier (I couldn't think of a sequence $\varepsilon_k \in o(2^k/k)$ that would make it easy to estimate $\sum_{0 < k \le n} 2^k/k - \varepsilon_k$)

*... and a few others, to no avail


Any hints?

An asymptotic expansion for large $n$ may also be derived as follows:
\begin{align*}
\sum\limits_{k = 1}^n {\frac{{2^k }}{k}} & = \frac{{2^n }}{n}\sum\limits_{k = 0}^{n - 1} {\frac{{2^{ - k} }}{{1 - k/n}}}  = \frac{{2^n }}{n}\sum\limits_{k = 0}^{n - 1} {2^{ - k} \sum\limits_{j = 0}^\infty  {\frac{{k^j }}{{n^j }}} }  \sim \frac{{2^n }}{n}\sum\limits_{j = 0}^\infty  {\frac{1}{{n^j }}\sum\limits_{k = 0}^\infty  {k^j 2^{ - k} } } \\ & = \frac{{2^{n + 1} }}{n}\bigg( 1 + \sum\limits_{j = 1}^\infty  {2^{j} A_j \!\left( {\tfrac{1}{2}} \right)\frac{1}{{n^j }}} \bigg)= \frac{{2^{n + 1} }}{n}\bigg( 1 + \sum\limits_{j = 1}^\infty  {\frac{A_j (2)}{{n^j }}} \bigg).
\end{align*}
Here $A_j$ denotes the $j$th Eulerian polynomial. In terms of the Stirling numbers of the second kind
$$
A_j (2) = \sum\limits_{k = 1}^j  k! \,S(j,k) .
$$
Thus,
$$\tag{1}
\sum\limits_{k = 1}^n {\frac{{2^k }}{k}}  \sim \frac{{2^{n + 1} }}{n}\left( {1 + \frac{1}{n} + \frac{3}{{n^2 }} + \frac{{13}}{{n^3 }} + \frac{{75}}{{n^4 }} + \frac{{541}}{{n^5 }} +  \ldots } \right)
$$
as $n\to +\infty$. The numbers $1,3,13,75,541,\ldots$ are the ordered Bell numbers. It follows from their asymptotic behaviour that the best approximation coming from $(1)$ is achieved by taking only $
\left\lfloor n\log 2 \right\rfloor+1$ terms in the asymptotic series.
