# Algorithmic Analysis – Show that \$ lg (n!) \$ Is or is not \$ o ( lg (n ^ n)) \$ and \$ omega ( lg (n ^ n)) \$

My instructor has assigned a problem that asks us to determine which asymptotic limits apply to a certain $$f (n)$$ for a certain $$g (n)$$, in my case $$f (n) = lg (n!)$$ and $$g (n) = lg (n ^ n)$$. For clarity, the convention we use in our class is: $$lg = log_2$$, the "binary logarithm".

I know that by approximation of Stirling, $$lg (n!)$$ grows up in $$O (n lg (n))$$and evaluate the limit $$lim_ {n to infty} frac {n lg (n)} {n lg (n)} = C$$, a constant> 0, etc. $$lg (n!)$$ is in $$theta ( lg n ^ n)$$.

$$theta$$ also means that my $$f (n)$$ is $$O (g (n))$$ and $$Omega (g (n))$$but that does not mean that my $$f (n)$$ is in $$o (g (n))$$ or $$omega (g (n))$$.

For that, I think I should evaluate $$lim_ {n to infty} slash { lg (n!)} { lg (n ^ n)}$$but I'm not sure.

What strategy would I use to show that $$f (n)$$ is in $$o (g (n))$$ or $$omega (g (n))$$? Would I evaluate $$lim_ {n to infty} slash { lg (n!)} { lg (n ^ n)}$$?