Let \(\{a_n\}\) be a sequence defined by \(a_1 = \sqrt{2}\) and \(a_{n+1} = \sqrt{2 + a_n}\) for \(n \geq 1\).
  1. Show that the sequence is increasing.
  2. Show that the sequence is bounded above.
  3. Prove that the sequence converges and find its limit.

Definitions and Theorems

Proof

(a) We first show that the sequence is increasing. To see this, we can use induction. For the base case, observe that \(a_1 = \sqrt{2} > 0\). Then

$$ \begin{align*} a_2 = \sqrt{2 + a_1} = \sqrt{2 + \sqrt{2}} > a_1 \end{align*} $$

For the inductive case, assume that any \(n \geq 1\) and that \(a_n > a_{n-1}\). We will show that \(a_{n+1} > a_n\). Then

$$ \begin{align*} a_{n+1} &= \sqrt{2 + a_{n}} \\ &> \sqrt{2 + a_{n-1}} \quad \text{(By Induction)} \\ &= a_{n} \end{align*} $$

Therefore, \(a_{n+1} > a_{n}\) for any \(n\). Thus, the sequence \(\{a_n\}\) is strictly increasing.

(b) Next, we want to show that the sequence is bounded above. We will show by induction that \(a_n < 2\) for any \(n\). For the base case, \(a_1 = \sqrt{2} < 2\). For the inductive case, suppose that \(a_n < 2\). We want to show that \(a_{n+1} < 2\)

$$ \begin{align*} a_{n+1} &= \sqrt{2 + a_{n}} \\ &< \sqrt{2 + 2} \quad \text{(By induction)} \\ &< 2 \end{align*} $$

Therefore, \(a_n < 2\) for any \(n\) as desired. This shows that \(\{a_n\}\) is bounded above.

(c) By (a) and (b), \(\{a_n\}\) is increasing and bounded above. Therefore, \(\{a_n\}\) is convergent by the Monotone Convergence Theorem. So there exists some \(L\) such that

$$ \begin{align*} \lim\limits_{n \rightarrow \infty} a_n &= L \\ \lim\limits_{n \rightarrow \infty} \sqrt{2 + a_{n-1}} &= L \\ \sqrt{2 + \lim\limits_{n \rightarrow \infty} a_{n-1}} &= L \quad \text{($\sqrt{}$ is continuous)}\\ \sqrt{2 + L} &= L \\ 2 + L &= L^2 \\ (L + 1)(L - 2) &= 0 \\ \end{align*} $$

Since \(a_n > 0\) for any \(n\), then \(L = 2\). \(\blacksquare\)

References