Looking for some advice on basic sequence convergence

Question

Hello I am trying to show for my analysis class the limit of the sequence $\displaystyle \frac{n}{4^n}$ is $0$. I am wanting to do such ideally making use of Bernoulli inequality

I will add what I have tried but I am having trouble putting it all together.

$$4^{n}=(1+3)^{n} \ge 1+3n \text{ for all } n \in \mathbb{N}$$

and so $$\frac{1}{4^{n}} \le \frac{1}{1+3n} \to 0 \text{ as } n \to \infty,$$ showing that $\frac{1}{4^n} \to 0$.

But now how can I deal with the $n$ in the numerator?

Alternatively restating above

$$2^{n}=(1+1)^{n} \ge 1+n \text{for all } n \in \mathbb{N}$$

and so $$\frac{1}{2^{n}} \le \frac{1}{1+n} \to 0$$ showing that $\frac{1}{2^n} \to 0$. Since $4^n \gt 2^{n}$ it follows that $$\frac{1}{4^{n}} \le \frac{1}{2^{n}}$$

So is this at all a right start? How could I incorporate the $n$ as I know I cant multiply limits because n itself is a divergent sequence.

Thanks

Answer 1

As you show: $$\frac{1}{2^n}\leq\frac{1}{n+1}$$ But $$\frac{1}{4^n}=\frac{1}{(2^2)^n}=\frac{1}{(2^n)^2}=\left(\frac{1}{2^n}\right)^2\leq\left(\frac{1}{n+1}\right)^2=\frac{1}{(n+1)^2}$$ Then $$\frac{n}{4^n}\leq\frac{n}{(n+1)^2}$$ Can you finish?

Answer 2

Apply L'Hospital rule.$$\lim_{n\to \infty}\frac{n}{4^n}=\lim_n \frac{1}{4^n\log 4}=0$$

Answer 3

You can use the same logic but apply another term from the binomial theorem to get the inequality $$ 4^n = (1+3)^n \ge 1 + 3n + 3^2 \binom{n}{2} = 1+ 3n + 9n(n-1)/2 $$ and now your bound becomes $$ \frac{n}{4^n} \le \frac{n}{1+ 3n + 9n(n-1)/2} \to 0 \text{ as } n \to \infty $$

Answer 4

One more way is to consider the series

$$ \sum_{k=1}^{\infty} k x^k $$

Show that it converges for $x=\frac{1}{4}$. If the series convergss, the summand tends to 0.