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

Question

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

Emma Aldridge · Accepted Answer

Another approach.

We will be using the assymptotic Stirling inequality formulas https://tutor.hix.ai #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#

Emilia Aguilar · Answer

See 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#

Prove that for #n &gt; 1# we have #1 xx 3 xx 5 xx 7 xx cdots xx(2n-1) &lt; n^n#?

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