Proof for sequence $\frac{n^5}{3^n} \rightarrow 0$

Question

I am improving my skill at formal sequence convergence proofs, I find them very tricky. I want to prove that:

$$\frac{n^5}{3^n} \rightarrow0$$

This should be read as "converges to zero", the question is, how large should $n$ be?

I would want a way to compare these two expressions I have trouble picking a big $n$ because I do not quite understand how to compare denominator and enumerator. Could someone drop a small hint so I can continue with my proof, I do not want to use any limit theorems. What I want is that for $n>n_0$:

$$\left| \frac{n^5}{3^n} \right|< \epsilon$$

Answer 1

Note that for $n \geqslant 6$, by binomial theorem, $$ 3^n = (1+2)^n = \cdots \geqslant ? $$

Answer 2

What you need to show is that for any $\varepsilon > 0$, there exists an $N$ so large that $n > N$ implies $n^5/3^n < \varepsilon$. Alternatively you can use l'hopitals rule.

Answer 3

You'd like $\dfrac{n^5}{3^n} < \epsilon$.

This is equivalent to $5 \log n - n \log 3 < \log \epsilon$.

Can you show that $5 \log n - n \log 3 \to -\infty$?

Answer 4

Consider the ratio of two successive terms,

$$\frac{t_{n+1}}{t_n}=\frac13\left(1+\frac1n\right)^5.$$

When

$$1+\frac1n<\sqrt[5]3$$ i.e. $n>4$, it becomes less than one, and from there the sequence decreases to $0$.

Answer 5

Let $e^y =3$ where $y >1.$

Then $\dfrac{n^5}{e^{(yn)}} \lt (\dfrac{6!}{y^6})\cdot \dfrac{1}{n}.$