Power Dominates Logarithm

Theorem
Let $\epsilon > 0$, $B \in \N$ be arbitrary.

Then there exists $N \in \N$ such that $\left({\ln n}\right)^B < n^\epsilon$ for all $n > N$.

Proof
First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$, i.e. set $B = 1$.

By Exponential is Strictly Increasing and Strictly Convex, the exponential is strictly increasing.

Therefore, $\ln n < n^\epsilon$ if and only if $n < \exp \left({n^\epsilon}\right)$.

Choose $k \in \N$ such that $k\epsilon > 2$ and $N = k!$.

By Taylor Series Expansion for Exponential Function we have:


 * $\displaystyle \exp \left({n^\epsilon}\right) = \sum_{m \mathop \ge 0} \frac{n^{m \epsilon}}{m!} > \frac{n^{k \epsilon}}{k!}$

Then we have for all $n > N$:

This completes the proof when $B = 1$.

Now let $\epsilon > 0$, $B \in \N$ be arbitrary.

By the above we can find $N \in \N$ such that $\displaystyle \log n < n^{\epsilon / B}$ for all $n > N$.

Then $\left({\ln n}\right)^B < \left({n^{\epsilon / B}}\right)^B = n^\epsilon$ for all $n > N$.

This completes the proof for arbitrary $B$.