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 $f(x)=\sum_{k=1}^n\frac{x^k}{k}$.  Then, we can write
$$f(x)=\sum_{k=1}^n \int_0^x t^{k-1}\,dk=\int_0^x \frac{1-t^n}{1-t}\,dt$$
so that $f(2)=\sum_{k=1}^n\frac{2^k}{k}=\int_0^2 \frac{1-t^n}{1-t}\,dt$.

Next, we choose $0<\delta<1$ and write
$$\begin{align}
\frac{n}{2^n}\int_0^2 \frac{1-t^n}{1-t}\,dt&=\frac{n}{2^n}\int_0^{2-\delta}\frac{1-t^n}{1-t}\,dt+\frac{n}{2^n}\int_{2-\delta}^2\frac{t^n-1}{t-1}\,dt \tag 1
\end{align}$$
We can apply the mean-value theorem for the second term on the right-hand side of $(1)$ to reveal that for some $\xi_n\in [2-\delta,2]$ 
$$\begin{align}
\frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt&=\frac{n}{2^n(\xi_n-1)}\int_{2-\delta}^2(t^n-1)\,dt\\\\
&=\frac{n}{2^n(\xi-1)}\left(\frac{2^{n+1}-(2-\delta)^{n+1}}{n+1}+\delta\right) \tag 2
\end{align}$$
Letting $n\to \infty$ in $(2)$ yields
$$\lim_{n\to \infty}\frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt=\frac{2}{\xi-1}$$
For any given $\epsilon>0$, we can choose $0<\delta<1$ so small that $\left|\frac{2}{\xi_n-1}-2\right|<\epsilon$.  Now, we proceed with that $0<\delta<1$ fixed.  
Noting that $\frac{1-t^n}{1-t}$ is positive and monotonically increasing on $[0,2]$, the first term on the right-hand side of $(1)$ is bounded above by $\frac{n}{2^n}\frac{(2-\delta)^n-1}{1-\delta}\to 0$ as $n\to \infty$.

Putting it all together, we obtain the coveted limit

$$\lim_{n\to \infty}\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}=2$$

