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

Note by Mishti Angel 3 years, 5 months ago

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It is false for n=1 – Shourya Pandey · 3 years, 5 months ago

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Its false for n=2 also – Arnab Bhattacharya · 2 years, 6 months ago

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 · 3 years, 5 months ago

@Djordje Marjanovic – It's true for n = 3, 4, 5, ... – Pi Han Goh · 3 years, 5 months ago

@Djordje Marjanovic – Maybe Mishit think \((n!)^2\geq n^n\) – Arbër Avdullahu · 3 years, 5 months ago

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 · 3 years, 5 months ago

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TopNewestIt is false for n=1 – Shourya Pandey · 3 years, 5 months ago

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Its false for n=2 also – Arnab Bhattacharya · 2 years, 6 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 · 3 years, 5 months ago

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– Pi Han Goh · 3 years, 5 months ago

It's true for n = 3, 4, 5, ...Log in to reply

– Arbër Avdullahu · 3 years, 5 months ago

Maybe Mishit think \((n!)^2\geq n^n\)Log in to reply

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 · 3 years, 5 months ago

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