Proving infinite limit $\frac{polynomial}{exponential}$, only using definition of limit

Question

I have problems finding a proof for limit of the sequence:

$(a_n) = \frac{n^4}{3^n}$ as $n \rightarrow \infty$

or

$\lim\limits_{x \to \infty} \frac{n^4}{3^n} = a^*$

Exponential functions grow faster than polynomials, so I know that $a^*$ should be 0.

I want to prove this by using the definition of limits:

$\forall_{\epsilon > 0}\exists_{n_{0}\in \mathbb{N}}\forall_{n\in \mathbb{N}}|a_n - a^*| < \epsilon$

I want to find $n_0$, so I have tried to isolate $n$ in:

$\frac{n^4}{3^n}-0 < \epsilon$

This resulted in $4log_3(n) - n < \epsilon$

I have also tried to find an intermediate function $f_n$ such that:

$\frac{n^4}{3^n}-0 < f_n < \epsilon$

I have tried $\frac{3^{n^4}}{3^n}$ and $n^4$ for $f_n$. The first one result in $n^4 -n < \epsilon$ and the second one results in $n < \sqrt[4]{\epsilon}$. None of these result in a function for the lowerbound of $n_0$ dependent on $\epsilon$.

Does anyone have some tips to help me out?

Answer 1

$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^4}{3n^4}=\frac13\left(1+\frac1n\right)^4.$$ For $n\ge10$ we have $$\left(1+\frac1n\right)^4\le1.1^4=1.4641\lt\frac32.$$ Hence for $n\ge10$ we have $$a_{n+1}\le\frac12a_n$$ and by induction $$a_n\le\frac{a_{10}}{2^{n-10}}.$$ Let $\varepsilon\gt0$ be given. Choose $N\ge10$ so that $$2^N\gt\frac{2^{10}a_{10}}\varepsilon.$$ Then for $n\ge N$ we have $$2^n\gt\frac{2^{10}a_{10}}\varepsilon$$ and so $$a_n\le\frac{a_{10}}{2^{n-10}}\lt\varepsilon.$$

Answer 2

hint

Let

$$b_n=\ln (a_n)=4\ln (n)-n\ln (3) $$

$$=n\Bigl(4\frac {\ln (n)}{n}-\ln (3)\Bigr) $$

use the fact that $(b_n) $ goes to $-\infty $ and you will find your $n_0$.

Answer 3

First of all note that the statement "exponential functions grow faster than polynomials" is an informal way of saying that the following equation holds $$\lim_{n\to\infty} \frac{n^{k}} {a^{n}} =0,\,k>0,a>1\tag{1}$$ The simplest proof of $(1)$ is via binomial theorem. Let $p$ be a positive integer greater than $k+1$ and $b=a-1>0$. Then from binomial theorem we have for $n>p$ $$a^{n} =(1+b) ^{n} =\sum_{i=0}^{n}\binom{n}{i}b^{i}>\binom{n}{p}b^{p}$$ Note that the expression $\binom{n} {p} $ is a polynomial in $n$ of degree $p$ with leading coefficient $1/p!$ and we thus have for $n>p$ $$0<\frac{n^{k}}{a^{n}}<\dfrac{n^{k}}{{\displaystyle \binom{n}{p}} }b^{-p}$$ The first fraction on the right is the ratio of two polynomials in $n$ and the degree of denominator is greater than that of numerator so the fraction tends to $0$. The second factor $b^{-p} $ is constant so that the overall expression tends to $0$ and hence $n^{k} /a^{n} $ tends to $0$.

The above argument can be easily converted into a proof based on definition of limit but the process is boring and I will only provide some hints. First show that there is a positive integer $n_{1}>p$ such that $$\binom{n} {p} >\frac{n^{p}} {2\cdot p!}, \forall n>n_{1}$$ This combined with the previous inequality leads us to $$0<\frac{n^{k}}{a^{n}} <\frac{2\cdot p!} {n^{p-k}} b^{-p} \leq\frac {2\cdot p!} {n} b^{-p} $$ and you can easily make the rightmost expression less than $\epsilon$ by choosing suitable $n$.