Proof strategy for $\lim_{n\to\infty} \frac{2^n}{3^{n+1}}$ In order to take the limit
$$\lim_{n\to\infty} \frac{2^n}{3^{n+1}}$$
I did:
$$3^{n+1}>n2^n\implies \frac{1}{3^{n+1}}<\frac{1}{n2^n}\implies\\\frac{2^n}{3^{n+1}}<\frac{2^n}{n2^n}\implies \frac{2^n}{3^{n+1}}<\frac{1}{n}\implies\\ 0 <\lim_{n\to\infty}\frac{2^n}{3^{n+1}}<\lim_{n\to\infty}\frac{1}{n} = 0$$ therefore by the squeeze theorem:
$$\lim_{n\to\infty}\frac{2^n}{3^{n+1}} = 0$$
Is my reasoning right? I only conjectured that $3^{n+1}>n2^n$ but I think there may be another way to prove it without induction. Any ideas?
 A: Just write 
$$\frac{2^n}{3^{n+1}} = \frac{1}{3}\cdot \left(\frac{2}{3}\right)^n$$
and note $(2/3)^n \to 0$ as $n \to \infty$ as $|2/3| < 1$ (recall, $x^n \to 0$ if $|x| < 1$).
A: Let $0<k<1$ and set $1+x=1/k$. Then we have Bernoulli's inequality
$$
(1+x)^n\ge 1+nx \quad\text{(for integer $n\ge0$ and real $x\ge-1$)}
$$
or
$$
\frac{1}{k^n}\ge 1+n\left(\frac{1}{k}-1\right)=\frac{k+n(1-k)}{k}
$$
so
$$
k^n\le\frac{k}{k+n(1-k)}
$$
By the squeeze theorem, $\lim_{n\to\infty}k^n=0$.
If $|k|<1$, then $-|k|\le k\le |k|$ so, again by the squeeze theorem, $\lim_{n\to\infty}k^n=0$.
Bernoulli's inequality can be easily proved by induction.
Note that this is usually crucial for proving that the geometric series converges, so the argument $x^n\to 0$ for $|x|<1$ because the geometric series converges can be circular, depending on how the convergence of the series is established.
In your case, direct induction is just easier, but less general.
By the way, the same inequality shows that, for $k>1$, $\lim_{n\to\infty}k^n=\infty$.
