Prove that for #n > 1# we have #1 xx 3 xx 5 xx 7 xx cdots xx(2n-1) < n^n#?
2 Answers
See explanation...
Explanation:
If
#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
#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#
Another approach.
Explanation:
We will be using the assymptotic Stirling inequality formulas
https://en.wikipedia.org/wiki/Stirling%27s_approximation
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Calling
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so for