$ a_1 = 1$
$ a_2 = \dfrac{a_1}{a_1.1 +1} = \dfrac{1}{1.1+1} = \dfrac{1}{2} = \dfrac{1}{2!}$
$ a_3 = \dfrac{a_2}{ 2. 1 +1 } = \dfrac{1}{2} :3 = \dfrac{1}{6} = \dfrac{1}{3!}$
$ a_4 = \dfrac{a_3}{3.1+1} = \dfrac{1}{6} : 4 = \dfrac{1}{24} = \dfrac{1}{4!}$
$ a_5 = \dfrac{a_4}{4.1+1} = \dfrac{1}{24} : 4 = \dfrac{1}{5!}$
$\to$ Ta có dạng tổng quát $a_n = \dfrac{1}{n!}$
$\to a_{65} = \dfrac{1}{65!}$
