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