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

$\underset{n\to \mathrm{\infty }}{\lim }n\sin (2\pi n!e)=\underset{n\to \mathrm{\infty }}{\lim }n\sin (\frac{2\pi }{n+1}+\frac{2\pi }{(n+1)(n+2)}+...).$

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

$\underset{n\to \mathrm{\infty }}{\lim }n\sin (\frac{2\pi }{n+1})=\underset{n\to \mathrm{\infty }}{\lim }n\frac{2\pi }{n+1}=2\pi$.