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

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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!

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

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 $$n\to\infty$$.
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 $$n \to \infty$$.