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

2 Answers
Sep 9, 2016

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#

Sep 10, 2016

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#