$\lim\limits_{n \to \infty} \frac{n 2^{n}}{3^{n}}$. without L'Hôpital's rule

Question

I have the following limit for which I cannot use L'Hôpital's rule:

$$\lim\limits_{n \to \infty} \frac{n 2^{n}}{3^{n}}$$

The only thing that I know is that $\lim\limits_{x \to \infty}\, \frac{x^{\alpha}}{a^{x}} = 0$ when $x > 0$ and $a > 1$.

I'm aware that the limit is 0, but I can't figure out how to calculate that, I tried to squeeze theorem but I'm stuck with: $$0 < \frac{n 2^{n}}{3^{n}} < $$

and I can't find another term that goes to zero that is bigger than the term in the middle.

I understand the question can be trivial but I've no problem with other limits. Thanks in advance for your insights.

Answer 1

The only thing that I know is that $\lim_{x \to \infty}\, \dfrac{x^{\alpha}}{a^{x}} = 0$ when $x > 0$ and $a > 1$.

Then use $$\frac{n2^n}{3^n} = \frac{n^1}{(3/2)^n}$$

Answer 2

Here is another way to approach it for the sake of curiosity.

According to the ratio test, one has that \begin{align*} \limsup_{n\to\infty}\frac{a_{n+1}}{a_{n}} = \limsup_{n\to\infty}\frac{2}{3}\left(\frac{n+1}{n}\right) = \frac{2}{3} < 1 \end{align*}

Consequently, the following series converges: \begin{align*} \sum_{n=1}^{\infty}\frac{n2^{n}}{3^{n}} = 1\times\frac{2}{3} + 2\times\frac{4}{9} + 3\times\frac{8}{27} + \ldots \end{align*}

By its turn, this means that its general term $a_{n}$ converges to zero, and we are done.

Hopefully this helps!

Answer 3

Bernoulli's inequality states that $(1+x)^n \ge nx$ for $x > -1$ and $n \ge 1$. This is easily proved by induction. In particular, if $x > 0$ then $$\left( \frac 1{x+1} \right)^n \le \frac 1{nx}.$$

If you try to get a $\frac 23$ on the left you will take $x = \frac 12$ which gives you $n \left( \dfrac 23 \right)^n \le 2$. Not exactly what you wanted, but at least it shows the sequence is bounded.

But, if you do the same thing with a smaller value of $x$, say $x = \frac 13$, you obtain $$n \left( \dfrac 34 \right)^n \le 3$$ which leads immediately to $$n \left( \dfrac 23 \right)^n \le 3 \left( \dfrac 89 \right)^n$$ and now you can apply the squeeze theorem.

Answer 4

Using binomial theorem: $$\left(\frac{3}{2}\right)^n= \left(1+\frac{1}{2}\right)^n= \sum\limits_{k=0}^n\binom{n}{k}\left(\frac{1}{2}\right)^k> 1+\frac{n}{2}+\frac{n(n-1)}{2}\cdot\frac{1}{4}$$ then $$0<\frac{n2^n}{3^n}=\frac{n}{\left(\frac{3}{2}\right)^n}<\frac{n}{1+\frac{n}{2}+\frac{n(n-1)}{8}}=\frac{1}{\frac{1}{n}+\frac{1}{2}+\frac{n-1}{8}}\to0, n\to\infty$$