Prove $\displaystyle\lim_{n\to\infty}{n^\alpha}{r^n}=0\:(\vert{r}\vert\lt1,\:\alpha\in\mathbb{R})$

Question

I want to prove $\displaystyle\lim_{n\to\infty}{n^\alpha}{r^n}=0\:(\vert{r}\vert\lt1,\:\alpha\in\mathbb{R})$.
If $\displaystyle\alpha=1$, I can prove the proposition as follows.

<Solution>
Let $\displaystyle\frac{1}{\vert{r}\vert}=1+h$, then $\displaystyle{h}\gt0$.
For sufficiently large $\displaystyle{n}$, $\displaystyle\frac{1}{{\vert{r}\vert}^n}=(1+h)^n=1+nh+\frac{n(n-1)}{2!}h^2+\cdots+h^n\gt\frac{n(n-1)}{2}{h^2}$
As a result, $\displaystyle{n{\vert{r}\vert}^n}\lt\frac{2}{(n-1){h^2}}$ and $\displaystyle\lim_{n\to\infty}\frac{2}{(n-1){h^2}}=0$.
Therefore, $\displaystyle\lim_{n\to\infty}{n^{}}{r^n}=0$.

I don't know how to expand from $\displaystyle\{1\}$ to $\displaystyle\{\alpha\mid\alpha\in\mathbb{R}\}$.
I would appreciate it if you could tell me the solution.

Answer 1

Consider the function $f(x)=x^\alpha|r|^x$ for $x>0$

Notice $f'(x)=\underbrace{\dfrac{f(x)}{x}}_{>0}\underbrace{(\alpha+x\ln(|r|))}_{<0}<0\quad$ for $x$ large enough.

This is because $|r|<1$ so $\ln|r|<0$ and $\alpha$ is a fixed number, so for $x\gg 1$ large enough, the quantity is negative.

We have $f(x)\ge 0$ and $f\searrow$ at infinity and since $f$ is continuous, then it is bounded on $\mathbb R^+$.

We can now express $0\le f(n)=n^\alpha|r|^n=2^\alpha(\frac n2)^\alpha |r|^\frac n2|r|^\frac n2=\overbrace{2^\alpha}^\text{fixed value} \underbrace{f(\frac n2)}_\text{bounded}|r|^\frac n2<cst\cdot|r|^\frac n2\to 0$