On proving a limit tending to zero

Question

Consider: $(1-p)^{n^{4/3}}$.

$p\in (0,1),\,p=f(n),\,pn\geq m\,\forall\, n\in\mathbb{N},\,m>0$

Does $\lim\limits_{n\rightarrow \infty} ((1-p)^{n^{4/3}})=0$?

I started writing out a series expansion as $|p|<1$:

$1-pn^{4/3}+\cdots$

and bounding the limit by the new limit:

$\lim\limits_{n\rightarrow \infty} ((1-\frac{m}{n})^{n^{4/3}})$

This approach does not end well.

Answer 1

$(1-p)^{n^\frac43} = (1-p)^{n\times n^\frac13} = \left((1-p)^n\right)^{n^\frac13}$. With $p \ge \frac{m}n$ we get $(1-p)^{n^\frac43} \le \left((1-\frac{m}n)^n\right)^{n^\frac13}$.

We also have $\lim_{n \to \infty} (1-\frac{m}n)^n = e^{-m} < 1$. That means we can find some constant $B$ (independent from $n$) with $0 < B < 1$ that for some $N$ we have $(1-\frac{m}n)^n < B$ for all $n > N$.

So we finally get for $n > N: (1-p)^{n^\frac43} \le \left((1-\frac{m}n)^n\right)^{n^\frac13} < B^{(n^\frac13)}$. Since $B < 1$ and $\lim_{n \to \infty}n^\frac13 = \infty$, we have $\lim_{n \to \infty} B^{(n^\frac13)} = 0$, which finally proves $\lim_{n \to \infty}(1-p)^{n^\frac43}=0.$

Answer 2

Since $n \geq 0$ and $p<1$ $$ (1-p)^{n^{4/3}} \leq (1-p)^{n} $$ Therefore by the squeeze theorem

$$ \lim_{n \rightarrow \infty} (1-p)^{n^{4/3}} \leq \lim_{n \rightarrow \infty} (1-p)^{n} \leq \lim_{n \rightarrow \infty} \left( 1- \frac{m}{n}\right)^n =0 $$

Remarks: $1- p \leq 1- \frac{m}{n} < 1$. From where the claim follows.

Answer 3

Take $p(n)=\dfrac mn$, which fulfills the hypothesis and is the worst case (slowest decrease of $1-p(n)$).

$$\lim_{n\to\infty}\left(1-\frac mn\right)^{n^{4/3}}=\lim_{n\to\infty}\left(\left(1-\frac m{n^3}\right)^{n^3}\right)^n.$$

The limit of the inner expression is $e^{-m}<1$ so that as of some $N$ the expression is bounded by $1-\epsilon$ and the $n^{th}$ power converges to $0$.