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

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