25.40. Does the sequence $a_{n}=\sin (2 \pi n!e)$ converge?
25.40. A n s w e r: yes, converges to 0. According to problem 25.39, the fractional part of the number $n!e$ is between 0 and $1 / n$. Therefore, $0<\sin (2 \pi n!e)<$ $<\sin (2 \pi / n)$ for $n \geqslant 4$.