# Sequence of Powers of Number less than One/Necessary Condition

## Theorem

Let $x \in \R$ be such that $\size{x} < 1$.

Let $\sequence {x_n}$ be the sequence in $\R$ defined as $x_n = x^n$.

Then $\sequence {x_n}$ is a null sequence.

## Proof 1

Without loss of generality, assume that $x \ne 0$.

Observe that by hypothesis:

$0 < \size x < 1$

Thus by Ordering of Reciprocals:

$\size x^{-1} > 1$

Define:

$h = \size x^{-1} - 1 > 0$

Then:

$x = \dfrac 1 {1 + h}$

By the binomial theorem, we have that:

$\left({1 + h}\right)^n = 1 + n h + \cdots + h^n > n h$

because $h > 0$.

By Absolute Value Function is Completely Multiplicative, it follows that:

$0 \le \size {x^n} = \size x^n = \dfrac 1 {\paren {1 + h}^n} < \dfrac 1 {n h}$
$\dfrac 1 n \to 0$ as $n \to \infty$
$\dfrac 1 {n h} \to 0$ as $n \to \infty$
$\size {x^n} \to 0$

as $n \to \infty$.

Hence the result, by the definition of a limit.

$\blacksquare$

## Proof 2

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

Suppose that:

$\exists N \in \N: \size x^N < \epsilon$

Then the result follows by the definition of a limit, because:

$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

where either Absolute Value Function is Completely Multiplicative is applied.

It remains to consider the case that:

$\forall n \in \N: \size x^n \ge \epsilon$

By the Archimedean Principle, we can choose a natural number $M$ such that:

$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$

Then, by Sum of Geometric Progression:

$\displaystyle M \epsilon \le \sum_{k \mathop = 0}^{M - 1} \size x^k = \frac {1 - \size x^M} {1 - \size x} < M \epsilon$

Hence the result.

$\blacksquare$

## Proof 3

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the Archimedean Principle, there exists a natural number $M$ such that:

$M > \dfrac 1 {\paren {1 - \size x} \epsilon}$

By the Well-Ordering Principle, there exists a smallest natural number $m$ such that:

$\exists N \in \N: m > M \size x^N$

Note that:

$m - 1 \le M \size x^{N + 1}$

By elementary algebra, it follows that:

$1 - \dfrac 1 m = \dfrac {m - 1} m \le \size x < 1 - \dfrac 1 {M \epsilon}$

Hence, $m < M \epsilon$.

Therefore:

$\size x^N < \epsilon$

By Absolute Value Function is Completely Multiplicative, it follows that:

$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

Hence the result, by the definition of a limit.

$\blacksquare$

## Proof 4

Define:

$\displaystyle L = \inf_{n \mathop \in \N} \size x^n$

By the Continuum Property, such an $L$ exists in $\R$.

Clearly, $L \ge 0$.

Suppose that $L > 0$.

Then, by the definition of the infimum, we can choose $n \in \N$ such that $\size x^n < L \size x^{-1}$.

But then $\size x^{n + 1} < L$, which contradicts the definition of $L$.

Therefore, $L = 0$.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number.

By the definition of the infimum, there exists an $N \in \N$ such that $\size x^N < \epsilon$.

It follows that:

$\forall n \in \N: n \ge N \implies \size {x^n} = \size x^n \le \size x^N < \epsilon$

where either Absolute Value Function is Completely Multiplicative is applied.

Hence the result, by the definition of a limit.

$\blacksquare$

## Also known as

This result and Sequence of Powers of Reciprocals is Null Sequence are sometimes referred to as the basic null sequences.