The Convergence of (a^n) for 0 ≤ a ≤ 1

Let $0 \leq a \leq 1$ and consider the sequence $(a^n)$. Observe that if $a = 0$ then this sequence converges to $0$ and if $a = 1$ then this sequence converges to $1$. As we will see in the theorem below, if $0 < a < 1$ then this sequence will converge to $0$.

Theorem 1: Let $0 < a < 1$. Then $(a^n)$ converges to $0$.

• Proof: First observe that $|a^n| < 1$ for every $n \in \mathbb{N}$ and so $(a^n)$ is a bounded sequece.

• We now show that $(a^n)$ is a decreasing sequence. Observe that for all $n \in \mathbb{N}$ we have that:

(1)

\begin{align} \quad a^{n+1} - a^n = a^n(a - 1) \end{align}

• Since $a^n > 0$ and $a - 1 < 0$ we have that $a^n(a - 1) < 0$ and so:

(2)

\begin{align} \quad a^{n+1} - a^n < 0 \quad \Leftrightarrow \quad a^{n+1} < a^n \end{align}

• So $(a^n)$ is a decreasing sequence. By The Monotone Convergence Theorem we have that $(a^n)$ is a convergent sequence. Let $A$ be the limit of this sequence.

• Now observe that $(a^{2n})$ is a subsequence of $(a^n)$ and so this subsequence must also converge to $L$. However, also observe that:

(3)

\begin{align} \quad a^n \cdot a^n = a^{2n} \end{align}

• Taking the limit of both sides as $n \to \infty$ gives us that $L^2 = L$. So $L = 0$ or $L = 1$. We note that $L \neq 1$ since $(a^n)$ is a strictly decreasing sequence and $a^n < 1$ for every $n \in \mathbb{N}$. So $L = 0$. In other words, $(a^n)$ converges to $0$.