# Puzzles/Analytical Puzzles/Surprising Limit/Hint

Show that

$\lim_{n\rightarrow\infty}n\sin(2\pi n!e)=2\pi.$

$e=1+\frac{1}{2}+\frac{1}{3!}+\frac{1}{4!}+...$
• $sin$ is a periodic function with periodicity $2\pi$.