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Please help me providing a direction on how to prove this. I can prove by randomly finding value of n that makes n! greater than 5^n. But can someone help me prove mathematically.

Hanusri
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2 Answers2

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Use induction.

If 5^n < n! and 5 <= n+1, then 5^(n+1) == 5 * 5^n < (n+1) * n! == (n+1)!.

comingstorm
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  • How we can assume 5 <= n+1. n can be 1 or 2 or 3. Also if we assume 5 <= n+1 then 5^n will be <= n!. But we want to prove 5^n < n! – Hanusri Jan 30 '16 at 01:35
  • It's not true for `n<4` anyway. Start with whatever value you randomly find that does make `n! > 5^n`. Asymptotic analysis means you get to choose any starting value you want! – comingstorm Jan 30 '16 at 01:38
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5^n = o(n!) <=> if n close to ∞ limit 5^n/n! close to 0

prove:

set An = 5^n/n!, when n > 10, An = 5^n/n! < 1/(n-5)!
for An > 0 and limit 1/(n-5)! close to 0, then 0 <= limit An <= limit 1/(n-5)! <= 0
so limit 5^n/n! close to 0, and 5^n = o(n!)
Liam
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foolSunMa
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