Puzzles|Analytical puzzles | Surprising Limit|Hint

Show that

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

(calculable in head!)

Hint

• Use the series expansion of e:

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

• ${\displaystyle sin}$ is a periodic function with periodicity ${\displaystyle 2\pi }$.

Solution