Answer to Question #186156 in Calculus for Muhammad

Question

1. Use the ratio a_n+1/a_n to show that the given siquence {a_n} is strictly increasing or strictly decreasing.

{n^{n^2}/(2n)!}_n=1^+infinity

Answer 1

a_n = $\left(n^{\smash{n^2}}\right)$$/(2n)!$

a_n+1 = $\left((n+1)^{\smash{(n+1)^2}}\right)/(2(n+1))!$

a_n+1/a_n = $\left((n+1)^{\smash{(n+1)^2}}\right) / (2(n+1))!) /$($\left(n^{\smash{n^2}}\right)/(2n)!)$ =

= $((2n)! *\left((n+1)^{\smash{(n+1)^2}}\right)$ ) / ($\left(n^{\smash{n^2}}\right) * (2n+2)!)$ =

= $\left((n+1)^{\smash{n^2+2n+1}}\right) / ((2n+1)*(2n+2)*$$\left(n^{\smash{n^2}}\right))$ =

=$\left((n+1)^{\smash{n^2+2n}}\right)$ $/ (2*(2n+1)*$ $\left(n^{\smash{n^2}}\right))$ =

=($\left((n+1)^{\smash{n^2+2n-1}}\right) *(1+1/n)/(2*(2+1/n)*$$\left(n^{\smash{n^2}}\right)$)

Let's calculate the limit a_n+1/a_n

$\lim$ (n$\to$ +$\infty$) ($\left((n+1)^{\smash{n^2+2n-1}}\right)*(1+1/n)/2*(2+1/n)*$ $\left(n^{\smash{n^2}}\right)$ =

=$\lim$ (n$\to$ +$\infty$ ) ($\left((n+1)^{\smash{n^2+2n-1}}\right) /(4*$$\left(n^{\smash{n^2}}\right)$) = +$\infty$

Thats mean that ratio is increasing

Note: $(2n)! / (2n+2)! = 1/((2n+1)*(2n+2))$