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$$\lim_{n \to -\infty} log(n!)-\sqrt{n} > 0$$

or in other words:

$$\exists N\in \mathbb{N} \text{ s.t. } \forall n>N, log(n!)-\sqrt{n} > 0$$

I tried recursion:

Finding $m'$ s.t. $\sqrt{m'+1}-\sqrt{m'}\leq log(m'+1)$ and $m\geq m' $ s.t. $log(m!)>\sqrt{(m)}$

Therefore we know it is true for $m$ and want to show it is true for $m+1$. We have $log((m+1)!)=log(m+1)+log(m!)$ and $\sqrt{m+1}=(\sqrt{m+1}-\sqrt{m})+\sqrt{m}$.

By condition we set and the property of $m$, we have it true for $m+1$. However, I'm struggling with the process of finding $m'$.

Thank you very much!

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    $\begingroup$ What is your question, what have you tried, and where are you stuck? $\endgroup$
    – Bonnaduck
    Sep 20, 2022 at 3:33
  • $\begingroup$ @Bonnaduck Sorry, just adjusted $\endgroup$
    – angushushu
    Sep 20, 2022 at 4:27
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    $\begingroup$ Please also change the title - you have an $n$ which I don't think is supposed to be there. $\endgroup$
    – ItsMe
    Sep 20, 2022 at 5:10
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    $\begingroup$ The limit is infinite, use Stirling's approximation. $\endgroup$
    – Sam
    Sep 20, 2022 at 9:24
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    $\begingroup$ @hardmath hey sorry for that, just corrected, changed all x's to n $\endgroup$
    – angushushu
    Oct 28, 2022 at 5:49

2 Answers 2

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For any $n$ you have $\log(n!)=\sum_{i=1}^n\log(i)\geq(n-1)\cdot\log(2)$. Hence, $$ \log(n!)-\sqrt{n}\geq(n-1)\cdot\log(2)-\sqrt{n}=\sqrt{n}(\sqrt{n}\cdot\log(2)-1)-\log(2), $$ where the right-hand side converges to $\infty$ as $n\to\infty$.

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    $\begingroup$ Hey sorry I can't accept 2 answers, but this is beautiful $\endgroup$
    – angushushu
    Sep 21, 2022 at 3:25
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Base case: $\log 6 = \log (3!) > \sqrt 3: 1.792 > 1.732$

$$ \begin{gather} n \ge 1 \implies \sqrt {n+1} - \sqrt{n} < 1 \\ n > e \implies \log n > 1 > \sqrt {n+1} - \sqrt{n} \\ \log ((n+1)!) - \log (n!) = \log n \implies \log ((n+1)!) - \log (n!) > \sqrt {n+1} - \sqrt{n} \end{gather} $$

For $n \ge e$, the LHS increases faster than the RHS as $n$ increases, so the inequality $\log (n!) > \sqrt n$ holds for all $n \ge 3$. Hence $\log (n!) - \sqrt n >0$ for all $n \ge 3$, including as $n \to \infty$.

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