Show that $\lim_{n\to\infty}\frac{n^p}{a^n}=0,p\in\mathbb{N},a>1$ [duplicate]

Question

Show that $$\lim_{n\to\infty}\dfrac{n^p}{a^n}=0,p\in\mathbb{N},a>1$$

I was really impressed when I understood that we can choose any $p$ and $a$ (satisfying $p\in\mathbb{N},a>1$) and the limit still holds.

I am not so sure about the proof though. I tried to apply the Stolz-Cesaro theorem: $$\lim_{n\to\infty}\dfrac{n^p}{a^n}=\lim_{n\to\infty}\dfrac{(n+1)^p-n^p}{a^{n+1}-a^n}=\lim_{n\to\infty}\dfrac{n^p+pn^{p-1}+\dots+1-n^p}{a^{n+1}-a^n}\\=\lim_{n\to\infty}\dfrac{pn^{p-1}+\dots+1}{a^{n+1}-a^n},$$ which isn't very useful. What's a better aproach? ( Without L'Hopital's rule. )

The limit when $a\in(0;1)$ is $+\infty$, right?

Answer 1

Here is a simple way: since $a^{-1} < 1$ simply note that $$\sum_{n=1}^\infty n^p a^{-n} < \infty$$ by the ratio test, which implies that $n^p a^{-n} \to 0$ as $n\to \infty$. This will also show that $n^p a^{-n} \to \infty$ if $a \leq 1$.

Answer 2

Note that

$$\frac{n^p}{a^n}=\left(\frac{n}{(\sqrt[p]{a})^n}\right)^p$$

where $\sqrt[p]{a}>1$ since $a>1$. Since $x^p$ is continuous for all $x$, this implies

$$\lim_{n\to\infty}\frac{n^p}{a^n}=\left(\lim_{n\to\infty}\frac{n}{b^n}\right)^p$$

where $b=\sqrt[p]{a}>1$. It is therefore sufficient to only consider the limit with $n$ in the numerator. Writing $b=1+w$, we have

$$0< \frac{n}{b^n}=\frac{n}{(1+w)^n}=\frac{n}{\sum_{k=0}^n\binom{n}{k}w^k}<\frac{n}{1+nw+\frac{n(n-1)}{2}w^2}<\frac{2}{(n-1)w}$$

We conclude

$$\lim_{n\to\infty}\frac{n^p}{a^n}=\left(\lim_{n\to\infty}\frac{n}{b^n}\right)^p=0^p=0$$

Answer 3

This follows from Binomial Theorem and Sandwich Rule. No transcendental function (e.g. $\ln$) nor advanced theorem is involved. Fix $p\in\mathbb{N}$. Let $b=a-1$, then $b>0$. For $n>p$, we have that \begin{eqnarray*} a^{n} & = & \left(1+b\right)^{n}\\ & = & 1+nb+\frac{n(n-1)}{2!}b^{2}+\ldots+\frac{n(n-1)\ldots(n-p)}{(p+1)!}b^{p+1}+\ldots+b^{n}\\ & \geq & \frac{n(n-1)\ldots(n-p)}{(p+1)!}b^{p+1}. \end{eqnarray*} Therefore, \begin{eqnarray*} 0 & \leq & \frac{n^{p}}{a^{n}}\\ & \leq & \frac{n^{p}}{\frac{n(n-1)\ldots(n-p)}{(p+1)!}b^{p+1}}\\ & = & \frac{(p+1)!}{b^{p+1}}\cdot\frac{1}{n}\cdot\frac{1}{(1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{p}{n})}\\ & \rightarrow & 0 \end{eqnarray*} as $n\rightarrow\infty$. By Sandwich Rule, the result follows.

Answer 4

Here is a simpler way to prove it :

Based on your assumptions the term $u_n = \frac{n^p}{a^n}$ is strictly positive. We can then employ the natural logarithm function $\ln$ as follow : $$ln(u_n) = p \ln(n) - n \ln(a) = -n \left(\ln(a) -p \frac{\ln(n)}{n} \right) $$ Using the fact that $\lim_{n\to \infty} \frac{\ln (n)}{n} = 0$ and $\ln(a) > 0$ we conclude that $\lim_{n\to \infty} \ln(u_n) = -\infty$ thus $\lim_{n\to \infty} u_n = 0$