Limit of $n^c/b^n$ when $n$ tends to infinity

Question

How do I prove that $\lim_{n\to\infty}\frac{n^c}{b^n}=0$, where $c\in\mathbb{R}$ and $b>1$? I've managed to prove it using L'Hospital rule working with $\frac{x^c}{b^x}$, but is there a simpler way using only sequences?

Answer 1

Another way to see it: $$ \sum_{n=1}^\infty \frac{n^c}{b^n} < \infty $$ by the ratio test, hence the sequence must converge to $0$.

edit: $$ \left| \frac{\frac{(n+1)^c}{b^{n+1}}}{\frac{n^c}{b^n}} \right| = \frac{(n+1)^c}{b^{n+1}} \cdot \frac{b^n}{n^c} = \left(\frac{n+1}{n}\right)^c \cdot \frac 1 b \to \frac{1}{b} < 1 $$ as $n\to\infty$.

Answer 2

Let $m$ be any integer greater than $c$ and $0$. Since $b-1>0$ we have for $n\geq m$: $$b^n=(b-1+1)^n=\sum_{k=0}^n\binom{n}{k}(b-1)^k\geq\binom{n}{m}(b-1)^m=:g(n)$$ Now the right side $g(n)$ is a polynomial in $n$ of degree $m>c$ in $n$. It follows that $n^c/g(n)\to0$ as $n\to\infty$ and therefore by the squeeze theorem: $$\lim_{n\to\infty}\frac{n^c}{b^n}=0$$