Example
Let \(x_1 = 2\) and let \(x_{n+1} = \frac{x_n + \frac{2}{x_n}}{2}\). Find \(\lim\limits_{n \rightarrow \infty} x_n\)

Strategy

If it’s not clear whether the sequence is increasing or decreasing, then sometimes it’s best to try and find the limit first assuming it exists and then work backwards. In this, we will see that the limit is \(\sqrt{2}\).

Proof

Recall the AM-GM inequality where for any \(a,b \in \mathbb{R}\)

$$ \begin{align*} \frac{a+b}{2} \geq \sqrt{ab} \end{align*} $$

So suppose \(a = x_n\) and \(b = \frac{2}{x_n}\). Then

$$ \begin{align*} \frac{x_n + \frac{2}{x_n}}{2} &\geq \sqrt{x_n \cdot \frac{2}{x_n}} \\ x_{n+1} &\geq \sqrt{2} \end{align*} $$

But we also that \(x_1 = 2 \geq \sqrt{2}\). Therefore, any subsequent term \((n \geq 2)\) is also greater than \(\sqrt{2}\). Thus, \(x_n \geq \sqrt{2}\). This sequence is also decreasing. To see this observe that

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

But notice now that \(x_n \geq \sqrt{2}\) so \(x_n^2 \geq 2\) Therefore, \(x_n^2 - 2 \geq 0\) for all \(n\). Then

$$ \begin{align*} x_n \geq x_{n+1} \end{align*} $$

Thus, \(x_n\) is decreasing. To find the limit, suppose that the limit is \(L\), then

$$ \begin{align*} \lim\limits_{n\to\infty} x_{n+1} &= \lim\limits_{n\to\infty} \frac{x_n + \frac{2}{x_n}}{2} \\ L &= \frac{L + \frac{2}{L}}{2} \\ L &= \frac{L^2 + 2}{2L} \\ 2L^2 &= L^2 + 2 \\ L^2 &= 2 \\ L &= \pm\sqrt{2} \\ \end{align*} $$

Thus, \(\lim\limits_{n \rightarrow \infty} x_n = \sqrt{2}. \ \blacksquare\)

References

  • Lecture Notes by Professor Chun Kit Lai