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.


$\blacksquare$

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