Let us consider:
X= 99√(99!), and:
Y= 100√(100!)
Determine, without direct evaluation and using analytic/semi-analytic (simple calculator + p&p) techniques, which of these is greater:
• (i) X or Y
• (ii) X/99 or Y/100
*** Computer programs/excel solver/advanced calculators/online solvers can be used only for purposes of verification.
Define f(n) = (n!)^(1/n). Then X=f(99) and Y=f(100)
99 and 100 are 'large' so Stirling's Approximation can be substituted in for n! without materially impacting the values. n! ~= (n/e)^n * sqrt(2*pi*n).
Then f(n) = (n!)^(1/n) ~= ((n/e)^n * sqrt(2*pi*n))^(1/n) = n/e * (2*pi*n)^(1/(2n))
Then X = f(99) ~= 99/e * (198*pi)^(1/198) = 37.6228
And Y = f(100) ~= 100/e * (200*pi)^(1/200) = 37.9924
So the answer to part (i) is Y is greater than X.
Also, X/99 = 0.380028 and Y/100 = 0.379924
So the answer to part (ii) is X/99 is greater than Y/100.
Note: No advanced solvers were used; only a common scientific calculator was used.
Solution
