Last modified on 30 January 2008, at 12:51

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.