Sequence of Powers of Number less than One/Sufficient Condition

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x \in \R$.

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

Let $\sequence {x_n}$ be a null sequence.


Then $\size x < 1$.


Proof

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.

$\blacksquare$