Answer to Question #205807 in Calculus for abcdefna

Question

Given a sequence a_n=((n!)²)/(2n)! . Evaluate lim n approaches infinity a_n

Answer 1

$\text{By ratio test, }\\ If \space lim_{n→∞}\frac{a_{n+1}}{a_n}=L\space and\space L<1 \space then \space lim_{n→∞}a_n=0 .\\ Here, \\ a_n=\frac{(n!)^2}{(2n)!}\\ lim_{n→∞}\frac{a_{n+1}}{a_n}=lim_{n→∞}\frac{\frac{((n+1)!)^2}{(2n+2)!}}{\frac{(n!)^2}{(2n)!}}\\ =lim_{n→∞}\frac{((n+1)!)^2(2n!) }{(2n+2)!(n!)^2}\\ =lim_{n→∞}\frac{((n+1).n!)^2(2n!) }{(2n+2).(2n+1).(2n)!(n!)^2}\\ =lim_{n→∞}\frac{(n+1)^2 }{(2n+2).(2n+1)}\\ \text{Taking }n^2\text{ common from numerator and denominator.}\\ =lim_{n→∞}\frac{(1+\frac{1}{n})^2 }{(2+\frac{2}{n}).(2+\frac{1}{n})}\\ =\frac{1}{4}$