Inequalities

prove that (n!)^2 > n^n

Note by Mishti Angel
5 years, 1 month ago

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It is false for n=1

Shourya Pandey - 5 years, 1 month ago

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Its false for n=2 also

Arnab Bhattacharya - 4 years, 2 months ago

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proof: let's take n=1 n! = 1! = 1 (n!)^2 = 1^2 = 1 n^n = 1^1 = 1 then 1 > 1 is false so this expression is false. /* for n=2 n! = 2 n^2 = 2^2 = 4 n^2 = 2^2 = 4 4 > 4 is false

Djordje Marjanovic - 5 years, 1 month ago

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It's true for n = 3, 4, 5, ...

Pi Han Goh - 5 years, 1 month ago

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Maybe Mishit think \((n!)^2\geq n^n\)

Arbër Avdullahu - 5 years, 1 month ago

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the question tht i meant is posted by arber nd the inequality holds good for n belonging to natural numbers other than 1 and 2.... we need to prove it theoritically

Mishti Angel - 5 years, 1 month ago

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