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

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 $n \to \infty$.
