Sequence of Powers of Number less than One

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Theorem

Let $x \in \R$.

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


Then:

$\size x < 1$ if and only if $\sequence {x_n}$ is a null sequence.


Rational Numbers

Let $x \in \Q$.

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


Then:

$\size x < 1$ if and only if $\sequence {x_n}$ is a null sequence.


Complex Numbers

Let $z \in \C$.

Let $\sequence {z_n}$ be the sequence in $\C$ defined as $z_n = z^n$.


Then:

$\size z < 1$ if and only if $\sequence {z_n}$ is a null sequence.


Normed Division Ring

Let $\struct {R, \norm {\,\cdot\,}}$ be a normed division ring

Let $x \in \R$.

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


Then:

$\norm x < 1$ if and only if $\sequence {x_n}$ is a null sequence.


Proof

Necessary Condition


[For other proofs of the Necessary Condition visit here.]


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:

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

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.

$\Box$


Sufficient Condition

By Reciprocal of Null Sequence:

$\sequence {x_n}$ converges to $0$ if and only if $\sequence {\dfrac 1 {x_n} }$ diverges to $\infty$.

By the definition of divergence to $\infty$:

$\exists N \in \N: \forall n \ge N: \size {\dfrac 1 {x_n} } > 1$

In particular:

$\size {\dfrac 1 {x_N} } > 1$

By Ordering of Reciprocals:

$\size {x_N} < 1$

That is:

$\size {x_N} = \size {x^N} = \size x^N < 1$


Aiming for a contradiction, suppose $\size x \ge 1$.

By Inequality of Product of Unequal Numbers:

$\size x^N \ge 1^N = 1$

This is a contradiction.


So $\size x < 1$ as required.

$\blacksquare$