Example
If \(|a| < 1\), then \(\lim\limits_{n \rightarrow \infty} a^n = 0\).

Proof

Assume that \(0 < a < 1\). Then, \(aa^{n} < 1 \cdot a^{n}\). Therefore

$$ \begin{align*} a^{n+1} < a^n. \end{align*} $$

Hence, this sequence is monotone decreasing. Furthermore, it is bounded below by \(0\) since \(a > 0\) and so \(a^n > 0\). Therefore, by the monotone convergence theorem, the limit exists. Let

$$ \begin{align*} L = \lim\limits_{n \rightarrow \infty} a^n. \end{align*} $$

To compute the limit, take the limit on both sides as follows

$$ \begin{align*} a^{n+1} &= aa^{n} \\ \lim\limits_{n \rightarrow \infty} a^{n+1} &= \lim\limits_{n \rightarrow \infty} aa^{n} \\ \lim\limits_{n \rightarrow \infty} a^{n+1} &= a\lim\limits_{n \rightarrow \infty} a^{n} \quad \text{(By the algebraic limit theorem)} \\ L &= aL \quad \text{(If $\lim\limits a^n = L$, then so is $\lim\limits a^{n+1} = L$)} \\ L - aL &= 0 \\ L(1 - a) &= 0 \end{align*} $$

But recall that \(a < 1\) so this means that \(1 - a > 0\). Therefore, we must have that \(L = 0\).

Now, suppose that \(-1 < a < 0\). Then write \(b = -a\) where \(0 < b < 1\). Then, we know that \(\lim\limits b^n = 0\) by the positive case. Furthermore, by the algebraic limit theorem with \(-1\) as a constant

$$ \begin{align*} \lim\limits_{n \rightarrow \infty} -b^{n} = (-1)\lim\limits_{n \rightarrow \infty} b^n = 0. \\ \end{align*} $$

But now observe that

$$ \begin{align*} a^{n} = (-b)^n = (-1)^nb^n \end{align*} $$

We know that \((-1)^n = \pm 1\). This means that

$$ \begin{align*} -b^n \leq a^n \leq b^n \end{align*} $$

By the squeeze theorem and since both \(\lim\limits_{n \rightarrow \infty} b^n = 0\) and \(\lim\limits_{n \rightarrow \infty} -b^n = 0\), then we must have that \(\lim\limits a^n = 0\) as desired. \(\ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai