$\lim\limits_{n \to \infty}\frac{n^{\alpha}}{c^n}=0.(\alpha>0,c>1)$ Problem
Assume that $\alpha>0,c>1.$ Prove $$\lim_{n \to \infty}\frac{n^{\alpha}}{c^n}=0.$$
Proof
Denote $b=c^{\frac{1}{\alpha}}$. Then $$\frac{n^{\alpha}}{c^n}=\frac{n^{\alpha}}{(b^{\alpha})^n}=\left(\frac{n}{b^n}\right)^{\alpha}.$$
Notice that $b=c^{\frac{1}{\alpha}}>1^{\frac{1}{\alpha}}=1$. We may assume that $b=1+h(h>0)$. Hence $$\forall n \geq 2:b^n=(1+h)^n=1+nh+\frac{n(n-1)}{2}h^2+\cdots \geq \frac{n(n-1)}{2}h^2,$$
Thus $$0 \leq \frac{n^{\alpha}}{c^n}=\left(\frac{n}{b^n}\right)^{\alpha}\leq \left(\frac{2}{(n-1)h^2}\right)^{\alpha} \to 0 (n \to \infty).$$
By the squeeze theorem, we may obtain $$\lim_{n \to \infty}\frac{n^{\alpha}}{c^n}=0.$$
 A: $c>1$. $c^n= \exp (n\log c)$, where $\log c >0.$
$\dfrac{n^a}{\exp (n\log c)}$;
Set $b:=\dfrac{a}{\log c} >0$.
$(\dfrac{n^b}{\exp n})^{\log c}.$
Take the limit $n \rightarrow \infty$.
A: If you assume
the binomial theorem,
you can argue directly
like this:
If
$n >m > a+1$
and
$c = 1+h$
 then
$\begin{array}\\
c^n
&=(1+h)^n\\
&=\sum_{k=0}^n \binom{n}{k}h^k\\
&>\binom{n}{m}h^{m}\\
&=\dfrac{\prod_{j=0}^{m-1}(n-j)}{m!}h^{m}\\
&=\dfrac{n^m\prod_{j=0}^{m-1}(1-j/n)}{m!}h^{m}\\
\text{so}\\
\dfrac{n^a}{c^n}
&\lt \dfrac{n^a}{\dfrac{n^m\prod_{j=0}^{m-1}(1-j/n)}{m!}h^{m}}\\
&\lt \dfrac{m!}{n^{m-a}h^m\prod_{j=0}^{m-1}(1-j/n)}\\
\end{array}
$
For fixed $m$ and $h$,
$\dfrac{m!}{h^m}
$
is fixed and
$\prod_{j=0}^{m-1}(1-j/n)
$
is an increasing function of $n$,
so
$\prod_{j=0}^{m-1}(1-j/n)
\gt \prod_{j=0}^{m-1}(1-j/(m+1))
$
so
$\dfrac{n^a}{c^n}
\lt \dfrac{g(m, h)}{n^{m-a}}
\lt \dfrac{g(m, h)}{n}
$
where
$g(m, h)$
is a function of $m$ and $h$,
so
$\dfrac{n^a}{c^n}
\to 0$.
A: Another Proof
First, we can prove that $$\forall k \in \mathbb{N}:\lim_{n \to \infty}\frac{n^{k}}{c^n}=0.$$
For this purpose, we assume that $c=1+h(h>0)$. Then $$\forall n \geq k+1:(1+h)^n\geq \frac{n(n-1)\cdots(n-k)}{(k+1)!}h^{k+1}.$$Thus
\begin{align*}
0 \leq \frac{n^k}{c^n} &\leq \frac{(k+1)!}{h^{k+1}}\cdot \frac{n^k}{n(n-1)\cdots(n-k)}\\&=\frac{(k+1)!}{h^{k+1}}\cdot \dfrac{1}{n\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{k}{n}\right)}\to 0(n \to \infty).
\end{align*}
By the squeeze theorem, we are done.
Now, it's obvious that we can always choose a fixed $k\in \mathbb{N}$ such that $\alpha\leq k.$ Then $$0 \leq \frac{n^\alpha}{c^n} \leq \frac{n^k}{c^n}\to 0(n \to \infty).$$
By the squeeze theorem, we obtain $$\lim_{n \to \infty}\frac{n^{\alpha}}{c^n}=0,$$which is what we want.
A: That nice, as an alternative without binomial theorem we can show that
$$\frac{n^{\alpha}}{c^n}=e^{\alpha \log n-n\log c} \to 0$$
indeed since $\frac{\log x}x\to 0 \implies \frac{\log n}n\to 0$ we have
$$\alpha \log n-n\log c = n\left(\alpha\frac{\log n}n-\log c\right) \to-\infty$$
and to prove $\frac{\log x}x\to 0$ assuming $x=e^y \to \infty$ since eventually $e^y>y^2$ we have
$$\frac{\log x}x=\frac{y}{e^y}<\frac1y\to 0$$
