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Puzzles|Analytical puzzles | Surprising Limit|Hint

Show that

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

(calculable in head!)

- Use the series expansion of e:

$e=1+{\frac {1}{2}}+{\frac {1}{3!}}+{\frac {1}{4!}}+...$

- $sin$ is a periodic function with periodicity $2\pi$.