5

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 $$\lim_{n \to \infty}\frac{n^{\alpha}}{c^n}=0.$$

Community
  • 1
mengdie1982
  • 13,840
  • 1
  • 14
  • 39

4 Answers4

1

$c>1$. $c^n= \exp (n\log c)$, 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 $n \rightarrow \infty$.

Peter Szilas
  • 20,344
  • 2
  • 17
  • 28
0

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 $\dfrac{n^a}{c^n} \to 0$.

marty cohen
  • 107,799
0

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.

mengdie1982
  • 13,840
  • 1
  • 14
  • 39
0

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

$$\frac{\log x}x=\frac{y}{e^y}<\frac1y\to 0$$

user
  • 154,566