What is
$\displaystyle
\lim_{n\to \infty} \frac{\pi}{2^n} \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})$
I'm getting 0 but I think i'm going wrong somewhere. Can anyone give me an idea?
What about this: $\displaystyle \sin\frac{j\pi}{2^n}$ is the imaginary part of $\displaystyle \left(\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}\right )^j$. Thus, if you make a detour through complex numbers, you can explicitly evaluate $\displaystyle \Sigma^{2^n}_{j=1} \sin(\frac{j\pi}{2^n})$, because you can then apply the formula for a geometric series.
No, I'm sorry but I get the answer 2. Quite possibly I am mistaken, therefore let me give you some of the my steps:
$\displaystyle \frac{\pi}{2^n}\sum_{k=1}^{2^n}\mathrm{e}^{\mathrm {i}\frac{k\pi}{2^n}}=\frac{\pi}{2^n}\left[\frac{\mathrm{e}^{\mathrm{i}\left(\pi+\frac{\pi}{2 ^n}\right)}-1}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}-1\right]$
$\displaystyle =\frac{\pi}{2^n}\left[\frac{-\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}-1\right]=-\frac{\pi}{2^n}\frac{2\mathrm{e}^{\mathrm{i}\frac{ \pi}{2^n}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2^n}}-1}$
$\displaystyle =-\frac{\pi}{2^n}\frac{2\mathrm{e}^{\mathrm{i}\frac{ \pi}{2^n}}}{\mathrm{i}\frac{\pi}{2^n}+o\left(\math rm{i}\frac{\pi}{2^n}\right)}\rightarrow 2\mathrm{i}$
where I have used Landau's little-o. Now this last expression seems to have the limit 2i, for $\displaystyle n\rightarrow\infty$, and the imaginary part of this would be 2.
P.S: I have confirmed that a numeric approximation quite rapidly converges to this solution.