Finding the limit of $a_n = \frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$

Assuming $a_n$ has a limit, which it appears to have, and your recursive expression is correct, which it appears to be, we can write $$\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{n+2}{2(n+1)}(a_n+1)$$ $$x=(x+1)\lim_{n\to\infty}\left(\frac{n+1}{2(n+1)}+\frac{1}{2(n+1)}\right)$$ $$x=\frac{x+1}{2}$$ $$x=1$$

OK, I see what needs to be done here. Denote the sum in brackets by $S_n$

$$S_n = \left(\frac{2}{1}+\frac{2^2}{2}+...+\frac{2^{n}}{n}\right)$$

First we prove by induction (for $n \ge 4$) the following:

$$S_n \gt \frac{ 2^{n+1} (n+2)}{(n+1)n} \tag{1}$$

Then using (1) it is easy to prove by induction that:

(2) the sequence is decreasing for $n\ge4$ i.e. $a_n \gt a_{n+1}$

(3) $a_n \ge 1$ for $n\ge4$ (that is even true for every $n$ but for us it's enough that it's true for $n\ge4$)

Once these two statements are proved, the convergence follows.
And then you take limits (that's the easy part as K.defaoite showed) and you're done.

Here is how using (1) we can prove the sequence is decreasing: we have $a_n=\dfrac{n+1}{2^{n+1}}S_n$ and $a_{n+1}=\dfrac{n+2}{2^{n+2}}S_{n+1}$ $=\dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$. Now, the condition that $a_n\gt a_{n+1}$ is that $\dfrac{n+1}{2^{n+1}}S_n\gt \dfrac{n+2}{2^{n+2}}(S_n+\dfrac{2^{n+1}}{n+1})$; this is equivalent to $2(n+1)S_n\gt (n+2)(S_n+\dfrac{2^{n+1}}{n+1})$, which is equivalent to $nS_n\gt \dfrac{(n+2)2^{n+1}}{n+1}$ (subtract $(n+2)S_n$ from either side), and dividing this by $n$ gives (1).