Example
Let \(x_1 = 1\) and let \(x_{n+1} = \sqrt{2 + x_n}\). Show that \(\lim\limits_{n \rightarrow \infty} x_n\) exists and find the limit.

Proof

We first need to show that \((x_n)_n\) is monotone increasing and bounded. We can use induction to show that it’s bounded above by \(2\). For the base case, we know that \(x_1 = 1 \leq 2\). For the inductive case, suppose the statement is true for \(n\). We will show it’s true for \(n+1\). Then

$$ \begin{align*} x_{n+1} = \sqrt{2 + x_n} \leq \sqrt{2 + 2} = 2. \end{align*} $$

Hence, this sequence is bounded above by \(2\). Next, we want to show that it’s monotone increasing. Observe that

$$ \begin{align*} \frac{x_{n+1}}{x_n} = \frac{\sqrt{2 + x_n}}{x_n} = \sqrt{ \frac{2 + x_n}{x_n^2} } = \sqrt{ \frac{2}{x_n^2} + \frac{1}{x_n} }. \end{align*} $$

But recall that \(x_n\) is bounded above by \(2\). So

$$ \begin{align*} \frac{x_{n+1}}{x_n} = \sqrt{ \frac{2}{x_n^2} + \frac{1}{x_n} } \geq \sqrt{ \frac{2}{2^2} + \frac{1}{2} } = 1. \end{align*} $$

Thus, \(x_{n+1} \geq x_n\) and the sequence is monotone increasing. Therefore, we can use the Monotone Convergence Theorem to conclude that the limit exists. Then,

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

To compute the limit, take the limit on both sides of

$$ \begin{align*} \lim\limits_{n \rightarrow \infty} x_{n+1} &= \lim\limits_{n \rightarrow \infty} \sqrt{2 + x_n} \\ L &= \sqrt{2 + L} \\ L^2 &= 2 + L \\ L^2 - L - 2 &= 0 \\ (L - 2)(L + 1) &= 0.\\ \end{align*} $$

Then \(L = 2\) or \(L = -1\). Clearly, \(L \neq -1\). Therefore, we must have that \(L = 2\). \(\ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai