Question

Prove that for `n > 1` we have `1 xx 3 xx 5 xx 7 xx cdots xx(2n-1) < n^n`?

Answer 1

See explanation...

Explanation:

If `n > 1` is even, then we can consider the terms in pairs from the middle out:

`1 xx 3 xx 5 xx 7 xx ... xx (2n - 1)`

`= 1 xx 3 xx ... xx (n-1) xx (n+1) xx ... xx (2n - 3) xx (2n - 1)`

`=prod_(k=1)^(n/2) (n-(2k-1))(n+(2k-1))`

`=prod_(k=1)^(n/2) (n^2-(2k-1)^2)`

`< prod_(k=1)^(n/2) n^2 = n^n`

If `n > 1` is odd, then the middle term is `n` and the other terms can be considered in pairs from the middle out:

`1 xx 3 xx 5 xx 7 xx ... xx (2n - 1)`

`= 1 xx 3 xx ... xx (n-2) xx n xx (n+2) xx ... xx (2n-3) xx (2n - 1)`

`=n * prod_(k=1)^((n-1)/2) (n-2k)(n+2k)`

`=n * prod_(k=1)^((n-1)/2) (n^2-4k^2)`

`< n * prod_(k=1)^((n-1)/2) n^2 = n*n^(n-1) = n^n`

Answer 2

Another approach.

Explanation:

We will be using the assymptotic Stirling inequality formulas
https://en.wikipedia.org/wiki/Stirling%27s_approximation

`log_e (n-1)! < n log_e n - n < log_e n!`

(`log_e (n-1)! < n log_e n - n` is true for `n ge 8`)

or

`(n!)/n < n^n e^(-n) < n!`

Calling `P_(2n-1) = Pi_(k=1)^n(2k-1)` we have

`P_(2n-1)=((2n)!)/(2^n n!)` we have

`P_(2n-1) < ((2n)^(2n+1)e^(-2n))/(2^n n^n e^(-n))=2^(n+1)n^(n+1)e^(-n)=(2(2/e)^n n) n^n`

Here `(2(2/e)^n n) < 1` for `n ge 10`

so for `n ge 10` according to Stirling approximation

`P_(2n-1) < n^n`