Based off of asymptotic growth, why would this statement be true: lg^100(n)= o(n^0.01) ? Would it be because any log is less than any exponential?

Question

Based off of asymptotic growth, why would this statement be true: lg^100(n)= o(n^0.01)? Would it be because any log is less than any exponential?

Answer 1

Once you have $\dfrac{\ln(n)}{n} \to 0 $, $\dfrac{\ln^a(n)}{n^b} \to 0 $ for any $a, b > 0$.

From $\dfrac{\ln(n)}{n} \to 0 $, we have $\dfrac{\ln(f(n))}{f(n)} \to 0 $ for any strictly increasing unbounded positive $f$.

Since $\dfrac{\ln^a(n)}{n^b} =\left(\dfrac{\ln(n)}{n^{b/a}}\right)^a =\left(\dfrac{(a/b)\ln(n^{b/a})}{n^{b/a}}\right)^a =(a/b)^a\left(\dfrac{\ln(n^{b/a})}{n^{b/a}}\right)^a $ setting $f(n) = n^{b/a}$ gives us what we want.