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Thread: induction inequality

  1. #1
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    induction inequality




    I have to prove that $\displaystyle 2^n < n! for all n>4 $ using induction.

    So,um I started like this,

    $\displaystyle p(n) : 2^n < n !$
    so $\displaystyle p(n+1) : 2^(n+1) < (n+1)!$
    => $\displaystyle 2^n * 2 < (n+1) * n!$
    => $\displaystyle 2^n < (n+1)/2 * n!
    $

    so um idk what to do next.How do I prove that this prove holds for all n>4
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  2. #2
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    Quote Originally Posted by NidhiS View Post



    I have to prove that $\displaystyle 2^n < n! for all n>4 $ using induction.

    So,um I started like this,

    $\displaystyle p(n) : 2^n < n !$
    so $\displaystyle p(n+1) : 2^(n+1) < (n+1)!$
    => $\displaystyle 2^n * 2 < (n+1) * n!$
    => $\displaystyle 2^n < (n+1)/2 * n!
    $

    so um idk what to do next.How do I prove that this prove holds for all n>4
    Compare the factors at both sides:

    $\displaystyle 2^n * 2 < (n+1) * n!~\implies~\underbrace{2^n<n!}_{true}~\wedge~\und erbrace{2<n+1}_{true\ for\ n>3}$
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