Limit: $\lim_{n\to \infty} \frac{n^5}{3^n}$ I need help on a homework assignment. How to show that $\lim_{n\to\infty} \left(\dfrac{n^5}{3^n}\right) = 0$? We've been trying some things but we can't seem to find the answer.
 A: First show that $3^n>n^6$ for all $n\geq15$.
This can be done inductively.
If $3^n>n^6$ then $3\cdot3^{n}>3\cdot n^6$ which is greater than $(n+1)^6$, for $n>15$ (because $3>(1+\frac1n)^6$ for all $n>15$).
Therefore, eventually $0<\frac{n^5}{3^n}<\frac1n$. It follows that $\frac{n^5}{3^n}\xrightarrow[n\to +\infty]{}0$.
A: Here's an "elementary" proof that doesn't use any theorems such as l'Hôpital. Start off by writing $\frac{n^5}{3^n} = \frac{n^5}{2^n} \left(\frac{2}{3}\right)^n$. The term $\left(\frac{2}{3}\right)^n$ tends to zero as $n \rightarrow \infty$, so we would be done if we showed that the term $\frac{n^5}{2^n}$ is bounded as $n \rightarrow \infty$. It is clear that this term is always positive. The derivative of the function given by $f(n) = \frac{n^5}{2^n}$ is equal to $f'(n) = -2^{-n} n^4 (n \log(2) - 5)$. For $n > \frac{5}{\log 2}$, we have $f'(n) < 0$, so $f(n)$ is decreasing there. In particular, for all $n > \frac{5}{\log 2}$ we get $0 < f(n) \leq f\left(\frac{5}{\log 2}\right)$ and hence $0 < \frac{n^5}{3^n} \leq f\left(\frac{5}{\log 2}\right) \left(\frac{2}{3}\right)^n$. The right hand side tends to zero as $n \rightarrow \infty$ and we can conclude that $\lim_{n \rightarrow \infty} \frac{n^5}{3^n} = 0$.
A: Depending on what have you studied and what are you supposed to know, you might be able to use one of the following tricks (we put $a_n:=\frac{n^5}{a^n}$):


*

*Show that $\lim\limits_{n\to\infty} \frac{a_n}{a_{n-1}} < 1$. (This is one of the basic tricks: If this limit is $q<1$, then eventually $\frac{a_n}{a_{n-1}}<q+\varepsilon$ for some small $\varepsilon$, and $a_n < C (q+\varepsilon)^n$, now you can use the sandwich theorem.)

*Show that $\lim\limits_{n\to\infty} \log a_n = -\infty$. (Then $\lim\limits_{n\to\infty} a_n=0$ follows from the continuity of $\log a$ for $a>0$.)

*L'Hospital criterion: Since both $f(x):=x^5$ and $g(x):=3^x$ are 5 times differentiable and have limit $+\infty$ up to $4$th derivative, you have
$$\lim\limits_{n\to\infty} \frac{a_n}{a_{n-1}} = \lim\limits_{n\to\infty} \frac{f'(x)}{g'(x)} = \cdots = \lim\limits_{n\to\infty}\frac{f^{(5)}(x)}{g^{(5)}(x)}.$$
It remains to show that the $5$th derivative of $f$ is constant and the $5$th derivative of $g$ has limit $+\infty$.
A: \begin{align*}
\lim_{n\to\infty} \left(\frac{\frac{\log(n^5)}{\log(3)}}{\log_3({3^n})}\right)
&= \lim_{n\to\infty} \left(\!\frac{\,\frac{5\log(n)}{\log(3)}\,}{n}\!\right) \\
&= \frac{5}{\log(3)} \lim_{n\to\infty} \left(\frac{\log(n)}{n}\right) \\
&= 0
\end{align*}
A: Let $a_n = \frac{n^5}{3^n}$. Then clearly $a_n >0$ for all $n \in \mathbb{N}$. The ratio of two consecutive terms of the sequence is:
$$
   \frac{a_{n+1}}{a_n} = \frac{(n+1)^5}{3^{n+1}} \frac{3^n}{n^5} = \frac{1}{3} \left(1+\frac{1}{n}\right)^5
$$
For $n \geqslant 5$:
$$
  \frac{a_{n+1}}{a_n} = \frac13\left(1+\frac{1}{n}\right)^5 \leqslant \frac{1}{3} \left(1+\frac{1}{5}\right)^5 < 1
$$
Hence for $n\geqslant 5$ the sequence is decreasing, and since sequence elements are positive the sequence converges to this lower bound, i.e. zero.
A: Consider the function $f(x)=a^x x^5$ with $0<a<1$.
We have $f'(x)= x^4 a^x (x \log a+5)$ and so $f$ has a single extreme at $x=-\dfrac{5}{\log a}>0$.
This extreme must be a maximum $M$ because $f(x)\ge 0$ and $f(0)=0$.
Now take $a=2/3$. Then, for all $n$, $\dfrac{n^5}{3^n} \le \dfrac{M}{2^n} \to 0$.
The same argument proves that $\dfrac{n^k}{b^n}\to 0$ for all $k>0$ and $b>1$.
A: Can you complete this?
$n>32 \to n^5<2^n \to \frac{n^5}{3^n}<\frac{2^n}{3^n}$
$n>\log_{\frac32}(\epsilon) \to (\frac32)^n>\epsilon \to (\frac{2}{3})^n<\epsilon$
