# Power Dominates Logarithm

## Theorem

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

Let $B \in \N$ be arbitrary.

Then there exists $N \in \N$ such that:

$\forall n > N: \paren {\ln n}^B < n^\epsilon$

## 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 $n > N$,.

Therefore:

$\ln n < n^\epsilon \iff n < \map \exp {n^\epsilon}$

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

$\ds \map \exp {n^\epsilon} = \sum_{m \mathop \ge 0} \frac {n^{m \epsilon} } {m!} > \frac {n^{k \epsilon} } {k!}$

Then for all $n > N$:

 $\ds \map \exp {n^\epsilon}$ $>$ $\ds \frac {n^{k \epsilon} } {k!}$ $\ds$ $>$ $\ds \frac N {k!} n^{k \epsilon - 1}$ $\ds$ $>$ $\ds n$ because $k \epsilon > 2$ and $N = k!$

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:

$\forall n > N: \paren {\ln n}^B < \paren {n^{\epsilon / B} }^B = n^\epsilon$

Hence the result.

$\blacksquare$