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So lets say I have $$ n^{2} \le n! $$

For what positive integers is this not true? $n=2$ and $ 3$

Base case? $$n=4 \implies 16 \le 24 $$

What is the inductive hypothesis and how do I show the inductive proof? Thanks

entrelac
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Rickz0rz
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  • Assume that for some $k>4$, $k^2\le k!$. Then, show this is true for $(k+1)^2\le (k+1)!$ – Mark Viola Aug 19 '15 at 18:22
  • If we restrain $n$ to be as $n\geq 4,;n\in\mathbb{N}$, and if for $n_0=4$ you just showed that it holds, then if we assume that the inequality holds for some natural $k>4$, then it is sufficient to prove that for some natural $k+1$ the above expression also holds. The main point is that whenever we fix $k$, provided it's greater than $4$, and knowing that the inequality holds, therefore we can say that it's surely true for all $k\geq 4$. – TheVal Aug 19 '15 at 18:23

2 Answers2

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If $n^2<n!$ and $n\ge4$ then $$(n+1)^2=n^2+2n+1<n!+2n+1<n!+3n<n!+n!n=n!(n+1)=(n+1)!$$

Another way: $$(n+1)^2=n^2\left(1+\frac1n\right)^2<n!\cdot 4<n!(n+1)=(n+1)!$$

ajotatxe
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For, $ x_0 = 4 $, $ n^2 \le n! $ is true.

Now, given, $n^2 \le n!$,

$(n+1)^2 = n^2 + 2n + 1$,

and, $ n^2 \le n! \implies n^2(n+1)^2 \le (n+1)^2n!$

$\implies (n+1)^2 \le (\frac{1}{n^2} + \frac{1}{n}) * (n+1)!$

$n > 4 => \frac{1}{n^2} < \frac{1}{16} $ and $\frac{1}{n} < \frac{1}{4} $, So,

$ (\frac{1}{n^2} + \frac{1}{n}) \lt \frac5{16} < 1$

As, $ (n+1)^2 \le \frac5{16} * (n+1)! \implies (n+1)^2 \le (n+1)!$

Hasit Bhatt
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