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I'm having trouble proving the following by induction:

$\forall n \in \mathbb{N}, (n\ge2) \implies ((\sqrt{2})^n \le n!)$

So far I solved the base case for n = 2.

$((\sqrt{2})^2 \le 2!)$ is true.

However I'm having trouble determining the induction hypothesis and conclusion, and am just generally lost on how to proceed. Any suggestions?

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3 Answers 3

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The induction hypothesis would be $(\sqrt{2})^k \leq k!$. Now multiplying both sides by $\sqrt{2} > 0$ we have $$(\sqrt{2})^{k+1} \leq k! \sqrt{2} \leq k!(k+1) = (k+1)!$$

since $k+1 > \sqrt{2}$ for $k > 1$.

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If $m!\ge(\sqrt2)^m,$

$(m+1)!\ge(m+1)(\sqrt2)^m$

So, it is sufficient to establish $(m+1)(\sqrt2)^m\ge(\sqrt2)^{m+1}$

which holds true $\iff m+1\ge\sqrt2\iff m\ge?$

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You would have to use the induction step $(\sqrt{2})^{n-1}\leq (n-1)!$, so you can note in your proof that you can finish using this way to see it:

$$(\sqrt{2})^{n-1}\sqrt{2}=(\sqrt{2})^{n}\leq n!=(n-1)!n$$

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