Power Dominates Logarithm

Theorem
Let $\epsilon \in \R_{>0}$.

Let $B \in \N$ be arbitrary.

Then there exists $N \in \N$ such that:
 * $\forall n > N: \left({\ln n}\right)^B < n^\epsilon$

Proof
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.

First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$,.

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

Therefore:
 * $\ln n < n^\epsilon \iff n < \exp \left({n^\epsilon}\right)$

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

By Taylor Series Expansion for Exponential Function:


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

Then for all $n > N$:

This completes the proof when $B = 1$.

Now let $B \in \N$ be arbitrary.

By the above we can find $N \in \N$ such that:
 * $\forall n > N: \ln n < n^{\epsilon / B}$

Then:
 * $\forall n > N: \left({\ln n}\right)^B < \left({n^{\epsilon / B} }\right)^B = n^\epsilon$

Hence the result.