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 Function is Completely Multiplicative, it follows that:
- $0 \le \size {x^n} = \size x^n = \dfrac 1 {\paren {1 + h}^n} < \dfrac 1 {n h}$
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 Squeeze Theorem for Real Sequences: Corollary:
- $\size {x^n} \to 0$
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