謎題|分析謎題 | 令人驚訝的極限|提示
證明
lim n → ∞ n sin ( 2 π n ! e ) = 2 π . {\displaystyle \lim _{n\rightarrow \infty }n\sin(2\pi n!e)=2\pi .}
(可以在腦中計算!)
e = 1 + 1 2 + 1 3 ! + 1 4 ! + . . . {\displaystyle e=1+{\frac {1}{2}}+{\frac {1}{3!}}+{\frac {1}{4!}}+...}
解答