Let \(c\) be a real number with \(|c| < 1\). Using the monotone convergence theorem show that \(c^n \rightarrow 0\).

Definitions and Theorems


Proof

Consider \(\{|c^n|\}\). Then, since \(|c| < 1\), then sequence is bounded above by \(1\) since

$$ \begin{align*} |c^n| = |c|^n < 1 \end{align*} $$

It is also bounded below by \(0\) since \(|c| < 1\) so \(|c^n| > 0\) for all \(n\). Therefore

$$ \begin{align*} 0 < |c^n| < 1 \end{align*} $$

The sequence is also monotone decreasing since

$$ \begin{align*} |c^{n+1}| = |c| \cdot |c^n| < 1 \cdot |c^n| = |c^n| \end{align*} $$

since it’s bounded and monotone decreasing where the infimum is zero, then by the Monotone Convergence Theorem

$$ \begin{align*} \lim\limits_{n \rightarrow \infty} |c^n| = \inf \{|c^n|\} = 0 \end{align*} $$

Now, by Exercise from section 2.3, since (|c^n| \rightarrow 0), then (c^n \rightarrow 0) as desired. \(\blacksquare\)


References