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

Proof
, 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.