Answer
$\frac{3}{2},\frac{5}{3},1,\frac{9}{25},\frac{1}{11}$
Work Step by Step
$a_{n}=\frac{2n+1}{n!+1}$
Substitute $1,2,3,4,5$ to $n$:
$a_{1}=\frac{2(1)+1}{1!+1}=\frac{3}{2}$
$a_{2}=\frac{2(2)+1}{2!+1}=\frac{5}{3}$
$a_{3}=\frac{2(3)+1}{3!+1}=\frac{7}{7}=1$
$a_{4}=\frac{2(4)+1}{4!+1}=\frac{9}{25}$
$a_{5}=\frac{2(5)+1}{5!+1}=\frac{11}{121}=\frac{1}{11}$
