Show that $a_n=\frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+..+\frac{2^n}{n}\right)$ converges and find its limit Sequence: $$a_n=\frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+..+\frac{2^n}{n}\right)$$
My Attempt: I showed the sequence is increasing by considering the difference $a_{n+1}-a_n>0.$ But I am having trouble in showing that the sequence is bounded. I have managed to show that $$\frac{(n+1)(2^n-1)}{n\times 2^{n+1}}\leq a_n\leq \frac{(n+1)(2^n-1)}{2^{n+1}}<\frac{n+1}{2}.$$ 
How do I proceed?

$$ \lim_{n\to +\infty}\frac{\sum_{k=1}^{n}\frac{2^k}{k}}{\frac{2^{n+1}}{n+1}}\stackrel{\text{Cesàro-Stoltz}}{=}\lim_{n\to +\infty}\frac{\frac{2^{n+1}}{n+1}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\color{red}{1}, $$
no major mystery here.
