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 \(|c| < 1\), we must have
$$
\begin{align*}
|c^n| = |c|^n < 1
\end{align*}
$$
Therefore, the sequence is bounded above by \(1\). Moreover, since \(|c| > 0\), then
$$
\begin{align*}
|c^n| > 0
\end{align*}
$$
Therefore, the sequence is bounded below by \(0\). Furthermore, the sequence is also monotone decreasing since
$$
\begin{align*}
|c^{n+1}| = |c| \cdot |c^n| < 1 \cdot |c^n| = |c^n|
\end{align*}
$$
Thus, \(\{|c^n|\}\) is bounded and monotone decreasing. Therefore, by the Monotone Convergence Theorem, it must be convergent.
Next we want to find its limit. Recall from [2.3] Exercise 10 that if \(|c^n| \rightarrow 0\), then we know \(c^n \rightarrow 0\). So we just need to find \(\lim_{n \rightarrow \infty} |c^n|\). [TODO]
References
- Problem Statement Source: Aleph 0