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Math Help - Don't understand this For which nonnegative integers n is nē < n!? prove your answer?

  1. #1
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    Don't understand this For which nonnegative integers n is nē < n!? prove your answer?

    This problem been giving me fits;

    Q: For which nonnegative integers n is nē < n!? prove your answer?
    __________________________________________________ ______________

    I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.

    This is what I am stuck at?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Grillakis View Post
    This problem been giving me fits;

    Q: For which nonnegative integers n is nē < n!? prove your answer?
    __________________________________________________ ______________

    I know I have to use Basis Step, Inductive Hypothesis and Inductive Step. But when doing the basis step, if you let n > 1, then n^(2) is larger than n!, so how can the Basis hold up? I though Basis is when the LHS is equal to the RHS or it has to be true like 4 > 2.

    This is what I am stuck at?
    Consider the function:

    f(k)=k^2(k+1)-(k+1)^2=k^3-2k+1

    k \in \mathbb{R} , then:

    f'(k)=3k^2-2

    Now if k\ge 1,\ f'(k)>0 and so f(k) is increasing, and as f(1)=0,\ f(k)\ge0 for all k\ge 1

    Hence for all natural numbers k\ge 1 we have:

    f(k)=k^2(k+1)-(k+1)^2 \ge 0

    or:

    k^2(k+1)\ge (k+1)^2 \ \ \ \ \ \ \ \ \ \ \ ...(1)

    Now for the induction step of the proof assume that for some integer k>1 that k^2\le k!, then:

    k^2(k+1)\le (k+1)!

    but by (1) above:

    (k+1)^2 \le k^2(k+1)\le (k+1)!

    etc...

    CB
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  3. #3
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    I see...thanks CB
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