Puzzles/Analytical Puzzles/Surprising Limit/Solution

Puzzles|Analytical puzzles|Surprising Limit|Solution

• Expand the argument of $sin$

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

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

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

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

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