Answer
$\frac{\sqrt 3-2}{2}$
Work Step by Step
$\sum\limits_{k=1}^3(-1)^{k+1}sin \frac{\pi}{k}=(-1)^{1+1}sin \frac{\pi}{1}+(-1)^{2+1}sin \frac{\pi}{2}+(-1)^{3+1}sin \frac{\pi}{3}=(-1)^{2}sin \frac{\pi}{1}+(-1)^{3}sin \frac{\pi}{2}+(-1)^{4}sin \frac{\pi}{3}=sin \frac{\pi}{1}+-sin \frac{\pi}{2}+sin \frac{\pi}{3}=0-1+\frac{\sqrt 3}{2}=\frac{\sqrt 3-2}{2}$
