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.
 A: 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.
A: $$
\begin{align}
\lim_{n\to\infty}\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}
&=\lim_{n\to\infty}\sum_{k=0}^{n-1}\frac1{1-\frac kn}2^{-k}\tag{1}\\
&=\sum_{k=0}^\infty2^{-k}\tag{2}\\[6pt]
&=2\tag{3}
\end{align}
$$
Explanation:
$(1)$: substitute $k\mapsto n-k$
$(2)$: Dominated Convergence $\left(\frac1{1-\frac kn}\le k+1\text{ when }n\ge k+1\right)$
$(3)$: geometric series
A: Note that
$$\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = \frac{n}{2^n}\sum_{k=0}^{n-1} \frac{2^{n-k}}{n-k} = \sum_{k=0}^{n-1} \frac{1}{2^k} +  \sum_{k=1}^{n-1} \frac{k}{2^k(n-k)}.$$
The limit of the first sum on the RHS is $\sum_{k=0}^\infty 2^{-k} = 2$ and we can show  that the limit of the second sum is $0$  giving the result
$$\lim_{n \to \infty}\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}= 2.$$
To show that 
$$\lim_{n \to \infty} \sum_{k=1}^{n-1} \frac{k}{2^k(n-k)} = 0, $$
we have, since $2^k > k(k-1),$
$$0 < \sum_{k=1}^n \frac{k}{2^k(n-k)}= \frac{1}{2(n-1)}+ \sum_{k=2}^{n-1} \frac{k}{2^k(n-k)} \\ < \frac{1}{2(n-1)}+ \sum_{k=2}^{n-1} \frac{1}{(k-1)(n-k)} \\ =  \frac{1}{n-1} \left(\frac{1}{2} + \sum_{k=2}^{n-1}\frac{1}{k-1} +  \sum_{k=2}^{n-1}\frac{1}{n-k} \right) \\ \frac{1}{n-1} \left(\frac{1}{2} + 2\sum_{k=1}^{n-2}\frac{1}{k} \right)$$
Clearly the RHS converges to $0$ as $n \to \infty$ since a harmonic sum is of order $\log n$.
