Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ 
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?

Here are a few remarks:


*

*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$. 
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.

*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed 
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
 &\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$


This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).


Let
$$ S_n=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}. \tag{1}$$
Clearly
$$ S_n=\sum_{k=1}^n\frac{n}{k}\frac1{2^{n-k}}\ge \sum_{k=1}^n\frac1{2^{n-k}}=2-\frac{1}{2^n}. $$
Define
\begin{eqnarray}
f_n(x)=\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}\left(\frac12x\right)^k
\end{eqnarray}
and then $f_n(0)=0,f_n(2)=S_n$ and
$$ f_n'(x)=\frac{n}{2^n}\sum_{k=1}^nx^{k-1}=\frac{n}{2^n}\frac{1-x^n}{1-x}.$$
So
\begin{eqnarray}
S_n&=&\frac{n}{2^n}\int_0^2\frac{1-x^n}{1-x}dx\\
&=&\frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx.
\end{eqnarray}
Since
$$ \int_0^1\frac{1-x^n}{1-x}dx=\ln n+\gamma+o(1),$$
one 
$$ \frac{n}{2^n}\int_0^1\frac{1-x^n}{1-x}dx=o(1). $$
Noting that if $x\in[1,2]$, then $x-1\ge1$ and $x^n-1\le x^n$, one has
$$ S_n\le o(1)+\frac{n}{2^n}\int_1^2\frac{x^n-1}{x-1}dx\le o(1)+\frac{n}{2^n}\int_1^2x^ndx=o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).\tag{2}$$
From (1) and (2), one has
$$ 2-\frac{1}{2^n}\le S_n\le o(1)+\frac{n}{2^n}\frac{1}{n+1}(2^{n+1}-1).$$
Letting $n\to\infty$, one has
$$ \lim_{n\to\infty}S_n=2. $$
