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 2π 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

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