Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$ Show that as $n\rightarrow \infty,$ $\frac{n^4}{3^n}\rightarrow 0$ linearly with rate $\frac{1}{3}$
linearly means take $a_n=\frac{n^4}{3^n}$
$\lim_{n\rightarrow \infty} (a_n)^\frac{1}{n}$
$\lim_{n\rightarrow\infty } (\frac{n^4}{3^n})^\frac{1}{n}=\frac{1}{3}$
how to solve this problem

Since
$$\lim_{n\to\infty}\left(\frac{n^4}{3^n}\right)^\frac{1}{n} = \lim_{n\to\infty}\frac{\left(n^4\right)^\frac{1}{n}}{\left(3^n\right)^\frac{1}{n}}=\lim_{n\to\infty}\frac{n^\frac{4}{n}}{3},$$
it will suffice to show that
$$\lim_{n\to\infty}n^\frac{4}{n} = \left(\lim_{n\to\infty} n^\frac{1}{n}\right)^4 = 1.$$
Now let $n^{1/n} = (1+a)$. Then by the binomial theorem,
\begin{eqnarray}
n = (1+a)^n &=& \sum_{k=0}^n \binom{n}{k}a^k\\
            &=& 1+ na + \frac{n(n-1)}{2}a^2 + \cdots + a^n\\
            &>& \frac{n(n-1)}{2}a^2.
\end{eqnarray}
Thus $a^2 < \dfrac{2n}{n(n-1)} = \dfrac{2}{n-1}$, which implies $a < \sqrt{\frac{2}{n-1}}$. Therefore,
$$1 \le \lim_{n\to\infty} n^{1/n} < \lim_{n\to\infty}\left(1 + \sqrt{\frac{2}{n-1}}\right) = 1.$$
