Let $k\in\mathbb{N}$ and $a>1$. Show that $\lim_{n\to\infty} \frac{n^k}{a^n}=0$.

Question

I think what I need to do is find the value of $n$ where $n^k<a^n$. I know that this value occurs whenever $n>k\log_an$, however I don't understand how to interpret this result into a general $N$ to pick as a maximum for the sequence convergence. What am I missing here?

Answer 1

English is not my native language, so forgive me for some wrong expression.

First we assume that $ k=1 $, then we can write $ a $ as $ 1+b $ where $ b>0 $. Because \begin{align*} (1+b)^{n}=1+nb+\frac{n(n-1)}{2}b^{2}+\dots>\frac{n(n-1)}{2}b^{2}, \end{align*} Then we can have \begin{align*} 0<\frac{n^{k}}{a^{n}}=\frac{n^{k}}{(1+b)^{n}}\leqslant \frac{n^{k}}{n(n-1)b^{2}/2}<\frac{2}{(n-1)}\to 0, \qquad n\to\infty, \end{align*} which tells us $ n/a^{n}\to0\,(n\to\infty) $. At last, we have that $ a^{1/k}>1 $ for any $ k\in\mathbb{N} $, so let $ a'=a^{k} $ and we obtain \begin{align*} \frac{n^{k}}{a^{n}}=\left( \frac{n}{a^{1/k}} \right)^k. \end{align*} That is finite production of sequence that converges to $ 0 $, so \begin{align*} \lim_{n\to\infty}\frac{n^{k}}{a^{n}}=0. \end{align*}

Answer 2

You need to do more than find the value of $n$ for which $n^k/a^n < 1$. In order to show $\lim_{n\to\infty} n^k/a^n=0$, you need to show that for any $\epsilon>0$, there exists a natural number $N$ so $n>N$ implies $|n^k/a^n|<\epsilon$, which sice $n^k/a^n$ is positive is equivalent to $n^k/a^n<\epsilon$.

This is a very well-known result, but it can be a little tricky to prove from first principles. You cannot just solve the inequality $n^k<\epsilon a^n$ for $n$, at least not using elementary functions.

A clever trick is to use the binomial theorem on $a^n$. Specifically, letting $x=a-1$, then $$ a^n=(x+1)^n=1+\binom{n}1x+\dots+\binom{n}jx^j+\dots+\binom{n}nx^n $$ In particular, $$ a^n > \binom{n}{k+1}x^{k+1}>\frac{(n-k)^{k+1}}{(k+1)!}x^{k+1} $$ Intuitively, this shows that $a^n$ goes to infinity faster than $n^{k+1}$, which is enough to show that $n^k/a^n$ goes to zero. In terms of the $\epsilon-N$ proof, in order to find $N$ for which $n>N$ implies $$ \frac{n^k}{a^n}<\epsilon $$ it suffices to find $N$ for which $n>N$ implies $$ \frac{n^k}{\frac{(n-k)^{k+1}}{(k+1)!}x^{k+1}}<\epsilon $$ This expression no longer mixes polynomials and exponentials, so it should be easier to work with.

Answer 3

$a>1;$

Consider $\frac{x^k}{a^x}=\frac{x^k}{e^{x\log a}};$

$k-$fold application of Hospital yields

$\left (\frac{k!}{(\log a)^k}\right )\left(\frac{1}{a^x}\right).$

Take the limit $x \rightarrow \infty.$