Puzzles|Analytical puzzles|Surprising Limit|Solution

• Expand the argument of ${\displaystyle sin}$

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

Since sine is a periodic function, adding or subtracting multiples of ${\displaystyle 2\pi }$ can not change the result, thus the first few terms on the right hand side can be dropped (those first few where ${\displaystyle {\frac {n!}{z!}}}$ is a whole number thus n is greater or equal to z), and one is left with

${\displaystyle \lim _{n\rightarrow \infty }n\sin(2\pi n!e)=\lim _{n\rightarrow \infty }n\sin({\frac {2\pi }{n+1}}+{\frac {2\pi }{(n+1)(n+2)}}+...).}$

Since ${\displaystyle \sin(x)\approx x+...}$ for small ${\displaystyle x}$ (Taylor expansion of sine), the limit is

${\displaystyle \lim _{n\rightarrow \infty }n\sin({\frac {2\pi }{n+1}})=\lim _{n\rightarrow \infty }n{\frac {2\pi }{n+1}}=2\pi }$.