Puzzles/Analytical Puzzles/Surprising Limit/Solution

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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

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

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