# The limit of $n^{-p}$ for $p$ a natural number

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$

Next we discover what sequences such as $\displaystyle \frac{1}{n^2}$ and $\displaystyle \frac{1}{n^3}$ tend to. This will be very useful later when combined with other techniques.

Theorem
Let $p\in \N$. Then $\displaystyle \lim_{n\to \infty} n^{-p}\to 0$.

Proof
Suppose $\varepsilon >0$ is given. Then, $n>\varepsilon ^{-1/p}$ implies that $n^{-p}<\varepsilon$. Thus let $N$ be an integer greater than $\varepsilon ^{-1/p}$. For all $n>N$ we have $|n^{-p}-0|=n^{-p}\lt \varepsilon$ so $\displaystyle \lim_{n\to \infty} n^{-p}\to 0$.

Example
Consider the sequence $\displaystyle a_n=\frac{(-1)^n}{n^3}$. Then,
$\left| a_n \right| =\left| \frac{(-1)^n}{n^3} \right| =\frac{1}{n^3}\to 0.$
Thus, $\displaystyle a_n=\frac{(-1)^n}{n^3}\to 0$.

Remark
In the theorem we can change the assumption ‘$p\in \N$’ to ‘$p\in \R$ with $p>0$’.
For example, $\displaystyle \frac{1}{n^\pi }$ tends to $0$.

To see the difficulty consider what $2^\pi$ means? How do we multiply 2 by itself $\pi$ times? We know how to multiply $2$ by itself $3$ times and $4$ times but what about somewhere inbetween?

Thus, the general statement relies on defining $x^a$ where $a$ is a real number and $x>0$. You may have seen that one solution to this is to define $x^a$ (for $x>0$) to be $\exp(a\ln x)$ where $\exp$ and $\ln$ are the exponential and natural logarithm functions. These functions and $x^a$ are not defined rigorously yet. Hence, for the moment we shall restrict to the case of $p\in \N$.