# Sequence of Powers of Number less than One/Necessary Condition/Proof 1

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## 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

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:

- $\paren {1 + h}^n = 1 + n h + \cdots + h^n > n h$

because $h > 0$.

By Absolute Value of Product, it follows that:

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

From Sequence of Reciprocals is Null Sequence:

- $\dfrac 1 n \to 0$ as $n \to \infty$

By the Multiple Rule for Real Sequences:

- $\dfrac 1 {n h} \to 0$ as $n \to \infty$

By the Corollary to the Squeeze Theorem for Real Sequences:

- $\size {x^n} \to 0$

as $n \to \infty$.

Hence the result, by the definition of a limit.

$\blacksquare$

## Sources

- 1953: Walter Rudin:
*Principles of Mathematical Analysis*: $3.20 e$ - 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 4$: Convergent Sequences: $\S 4.12$: Example