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Show that, for any integer k > 2, if n ≥ k², then kⁿ < n!?
Where n is also an integer.
Ben, no, I mean k^n less than the factorial of n.
1 Answer
- nudnik0Lv 610 years agoFavourite answer
Since k > 2 and n ≥ k^2, n ≥ 9. It will be enough to show that n! > n^(n/2).
If n is even, n! is the product of T[j] := j(n + 1 - j) over 1 ≤ j ≤ n/2. For j=1, ..., n/2 - 1,
T[j] - T[j+1] = j(n + 1 - j) - (j + 1)(n - j) = 2j - n < 0
and therefore T[1] < T[2] < ... < T[n/2]. So,
n! > T[1]^(n/2) = n^(n/2).
If n is odd, n! = T[1]...T[(n - 1)/2] ((n + 1)/2), and we still have T[1] < ... < T[(n-1)/2], so
n! > T[1]^((n - 1)/2) ((n + 1)/2) = n^((n - 1)/2) ((n + 1)/2).
Since
(n + 1)/2 - √n = (√n - 1)^2/2 ≥ 0,
this implies that
n! > n^((n - 1)/2) √n = n^(n/2).
