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