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).


$\begin{array}\\
\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}
&=n\sum_{k=1}^n \frac{2^{k-n}}{k}\\
&=n\sum_{k=0}^{n-1} \frac{2^{-k}}{n-k}\\
&=\sum_{k=0}^{n-1} \frac{2^{-k}}{1-k/n}\\
\text{so}\\
\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2
&=\sum_{k=0}^{n-1} 2^{-k}(\frac1{1-k/n}-1)-\frac1{2^{n-1}}\\
&=\sum_{k=0}^{n-1} 2^{-k}(\frac{k/n}{1-k/n})-\frac1{2^{n-1}}\\
\end{array}
$
If $k \le cn$,
$2^{-k}(\frac{k/n}{1-k/n})
\le 2^{-k}\frac{c}{1-c}
$
so
$\sum_{k=0}^{\lfloor nc \rfloor} 2^{-k}(\frac{k/n}{1-k/n})
\le 2\frac{c}{1-c}
\lt 4c$
if
$0 < c < \frac12$.
If
$k > cn$,
$\begin{array}\\
\sum_{k=cn}^{n-1} 2^{-k}(\frac{k/n}{1-k/n})
&<\sum_{k=cn}^{n-1} 2^{-cn}(\frac{k/n}{1-k/n})\\
&=2^{-cn}\sum_{k=cn}^{n-1} (\frac{k}{n-k})\\
&<2^{-cn}\sum_{k=cn}^{n-1} n\\
&<n^22^{-cn}\\
&<e^{-cn \ln 2 + 2\ln n}\\
\end{array}
$
For any fixed $c > 0$,
$-cn \ln 2 + 2\ln n
\to -\infty$
as
$n \to \infty$,
so that
$e^{-cn \ln 2 + 2\ln n}
\to 0$.
By first choosing $c$ small
and then $n$ large,
both
$4c$ and
$e^{-cn \ln 2 + 2\ln n}$
can be made as small as we want,
so that
$\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}-2
$
can be made as small as we want,
so that
$\lim_{n \to \infty}\frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}
=2
$.
