Answer
$2, \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}$
Work Step by Step
Given: $a_{1}=2,a_{(n+1)}=\frac{a_{n}}{1+a_n}$
$a_{1}=6$
$a_{1+1}=a_{(2)}=\frac{a_{1}}{1+a_1}=\frac{2}{1+2}=\frac{2}{3}$
$a_{2+1}=a_{(3)}=\frac{a_{2}}{1+a_2}=\frac{2/3}{1+\frac{2}{3}}=\frac{2}{5}$
$a_{3+1}=a_{(4)}=\frac{a_{3}}{1+a_3}=\frac{2/5}{1+\frac{2}{5}}=\frac{2}{7}$
$a_{4+1}=a_{(5)}=\frac{a_{4}}{1+a_4}=\frac{2/7}{1+\frac{2}{7}}=\frac{2}{9}$
Hence we see that the first five terms are $2, \frac{2}{3},\frac{2}{5},\frac{2}{7},\frac{2}{9}$.
