Proving $ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$ I'm having trouble with a proof of this question:
$$ \lim_{n \to \infty} \, \left(\frac{2}{3}\right)^n = 0$$
I know $N(0)$ is a log of epsilon with base $\frac{2}{3}$ but I can't proceed with a proof.
 A: Well firstly to prove it we should use the $\epsilon-N$ definition of a limit.
Mathematical Working
(This is just working, not actual proof yet, just showing how to get to proof.)
Consider:
Let $n\in\mathbb{N}$
Trivially:
$|(\frac{2}{3})^n-0|=|(\frac{2}{3})^n|$
Now note that $3^n>n2^n$, proof of this inequality is as following:
Let $P_k$ be the statement that for some n=k:
$3^k>k2^k$
$P_1$: $3^1 = 3$, $(1)2^1 = 2$, since $3>2$, $P_1$ is true
Now let us assume $P_k$ is true and consider $P_{k+1}$
$3^{k+1}=3*3^k>3k2^k=\frac{3}{2}k2^{k+1}$
Now note:
$\frac{3}{2}k>k+1 \implies 3k > 2k + 1 \implies k>1$
Hence:
$\frac{3}{2}k2^{k+1}>(k+1)2^{k+1} \implies 3^{k+1}>(k+1)2^{k+1}$
Thus $3^n>n2^n, \forall n\in \mathbb{N}$
Hence $|\frac{2^n}{3^n}|<|\frac{2^n}{n2^n}|=|\frac{1}{n}|<\epsilon$
This occurs when $n>|\frac{1}{\epsilon}|=\frac{1}{\epsilon}$
So now with this here's the proof:
Let $N = \frac{1}{\epsilon}$ where $\epsilon>0$ is arbitrary but fixed.
If $n>N$ then $n>\frac{1}{\epsilon}$. This implies $\frac{1}{n}<\epsilon$
Since $n>1$, $n=|n|$, thus:
$|\frac{1}{n}|<\epsilon$
Now:
$|(\frac{2}{3})^n-0|<|\frac{2^n}{2^nn}|<|\frac{1}{n}|<\epsilon$
Thus given any $\epsilon > 0, \exists N$ s.t. $|(\frac{2}{3})^n-0|<\epsilon$ whenever $n>N$.
Thus $\lim_{n\rightarrow \infty}(\frac{2}{3})^n = 0$
A: In order to avoid using the logarithm, we shall use the following tactic.
Since $\frac 3 2=1+\frac 12$, using the binomial theorem, we know that
$$
\left ( \frac{3}{2}  \right ) ^n=\left ( 1+\frac{1}{2}  \right ) ^n=\sum_{k=0}^{n}\binom{n}{k}\left ( \frac{1}{2} \right )^k>\frac{n}{2}  .     $$
Therefore $\left(\frac 2 3\right)^n<\frac 2 n\to 0$.
A: The limit exists because the sequence is monotonic and bounded.
L'hopital:   $$c:=\lim_{n\to\infty}(\dfrac 23)^n=\lim_{n\to\infty}\dfrac {2^n}{3^n}=\lim_{n\to\infty}\dfrac {\ln2\cdot 2^n}{\ln3\cdot 3^n}=\dfrac {\ln2}{\ln3}\lim_{n\to\infty}\dfrac {2^n}{3^n}=\dfrac {\ln2}{\ln3}\lim_{n\to\infty}(\dfrac 23)^n=\dfrac {\ln2}{\ln3}c$$.
Thus $c=0$.
(You actually don't need L'hopital.   You can just split off a factor of $\frac 23$.)
A: Since $\{(\frac{2}{3})^n\}$ is strictly decreasing, we only need to show that its inf is $0$.
In fact you can show that if $0<q<1$ , then $$\inf\{q^n|n\in\mathbb{Z}_+\}=0$$. To prove this we assume the contrary, or $i=\inf\{q^n|n\in\mathbb{Z}_+\}>0$(since the inf of the set must exist and it cannot be negative) . By the definition
of inf, exists $n\in\mathbb{Z}_+$ subject to $$i<q^n<\frac{i}{q}$$, which implies $q^{n+1}<i$, a contradiction with the definition of inf.
A: A bit of an uncommon method:
By this question here, $$\frac{2}{3} = \lim_{n \to \infty}\big[\sum_{k=0}^{n}{\frac{(-1)^k}{2^k}}\big]$$
Alternatively, $$\sum_{k=0}^{n}{\frac{(-1)^k}{2^k}} = \frac{1}{3}\big(\big(-\frac{1}{2}\big)^n + 2\big) = \frac{\frac{2^{n + 1} + e^{i\pi n}}{2^n}}{3}$$
Hence, $$\lim_{n \to \infty}{\big(\frac{2}{3}\big)^n} \implies \lim_{n \to \infty}{\frac{\frac{(2^{n + 1} + e^{i\pi n})^n}{2^{n^2}}}{3^n}}$$
$$\to \lim_{n \to \infty}{\frac{\frac{1}{3^n}}{\frac{2^{n^2}}{(2^{n + 1} + e^{i\pi n})^n}}}$$.
$e^{i\pi n}$ is bounded and insignificant towards large $2^{n + 1}$ as $n \to \infty$.
$$\approx \lim_{n \to \infty}{\frac{\frac{1}{3^n}}{\frac{2^{n^2}}{(2^{n + 1})^n}}} \to 0$$
