A limit involving $\sum_{k=1}^{n}\frac{2^k}{k}$ I would like to show that
$\lim_{n \to \infty} \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} = 2$
I have seen proofs that $\sum_{k=1}^n \frac{2^k}{k} \sim \frac{2^{n+1}}{n}$ using the Euler-Maclaurin method but instead I tried with the squeeze theorem.
For one side I showed that
$$\frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{k} > \frac{n}{2^n} \sum_{k=1}^n \frac{2^k}{n} = \frac{1}{2^n}\sum_{k=1}^n 2^k = \frac{2^{n+1} - 2}{2^n} = 2 - \frac{1}{2^{n-1}}$$
and the limit on this side is $2$.  
Now I am having trouble finding an upper bound that converges to $2$.  
Thank you.

It is enough to apply the Stolz-Cesàro theorem. By setting $A_n=\sum_{k=1}^{n}\frac{2^k}{k}$ and $B_n=\frac{2^n}{n}$ we have
$$ \frac{A_{n}-A_{n-1}}{B_n-B_{n-1}} = \frac{\frac{2^n}{n}}{\frac{2^n}{n}-\frac{2^{n-1}}{n-1}} = \frac{1}{1-\frac{1}{2}\cdot\frac{n}{n-1}}\tag{1}$$
hence
$$ \lim_{n\to +\infty}\frac{A_{n}-A_{n-1}}{B_n-B_{n-1}}=2 \tag{2}$$
and
$$ \lim_{n\to +\infty}\frac{n A_n}{2^n} = 2 \tag{3} $$
as wanted.
