Question

Prove that `(n !)^2<n^nn !<(2n)!` for all positive integers `ndot`

Answer 1

We have,
`(n!)^(2)=(n!)=(1xx2xx3xx4xx.. Xx(n-1)n)(n!)`
Now, ` 1 le n, 2 le n, 3 le n, .., n le n`
or `1xx2xx3..(n-1)n le n xx n xx n..n`
or `n! le n^(n)`
or `(n!)(n!) le (n!)n^(n)`
or `(n!)^(2) le n^(n) (n!)`
Also, `(2n)!=1xx2..n xx (n+1)..(2n-1)xx(2n)`
Now, `n+1 gt n, n+2 gt n, n+3 gt n, .., n+n gt n`
or `(n+1)(n+2)(n+3)..(2n-1)(2n) gt n^(n)`
or ` n!(n+1)(n+2)..(2n-1)(2n) gt n!n^(n)`
or `(2n)! gt n! n^(n)`
`implies n! n^(n) lt (2n)!`
From (1) and (2), we get `(n!)^(2) le n^(n)(n!)lt (2n)!`