If $n^k<2^n$, then $c\cdot n^k<2^n$

Question

The following seems true, but I was wondering what the proof is.

I know that there exists $N\in\mathbb{N}$ such that $n^k<2^n$ for $n\in\mathbb{N}, n\geq N$. Is it true that there exists $N'\in\mathbb{N}$ such that $c\cdot n^k<2^n$ for $n>N'$ for $c>0$.

so to be clear: You have $c>0$ and want to find $N'\in\mathbb N$ such that $c\cdot n^k<2^n$ for all $n<N'$, correct? By the way, what is $k$? I assume it is a fixed positive integer. — Pink Panther, May 11 '19 at 10:25
Yes you are right, sorry I should have mentioned that, $k$ is a positive integer — , May 11 '19 at 10:27
This follows from $n^{-k} 2^n \to \infty$, or equivalently, $n^k 2^{-n} \to 0$. — Theo Bendit, May 11 '19 at 10:39

Maximilian Janisch · Accepted Answer · 2019-05-11T12:35:38.857

Your statement is correct:

It suffices to show this for natural numbers $k$ (since for a real number $r$ we have $n^{\lceil r\rceil}\geq n^r$.) Let $c$ be a fixed real number.

We have (by definition) \begin{equation} 2^n = \exp(n \cdot \ln 2) := \sum_{m=0}^\infty \frac{(n \ln 2)^m}{m!} > n^{k+1} \underbrace{\frac{(\ln 2)^{k+1}}{(k+1)!}}_{=:C>0} \end{equation} Thus $2^n > C\cdot n^{k+1} > c \cdot n^k$ for all $n > \dfrac c C$.

If $n^k<2^n$, then $c\cdot n^k<2^n$

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