Evaluating Limits of Recursive Sequences Examples 1

# Evaluating Limits of Recursive Sequences Examples 1

We will now look at some more examples of evaluating limits of recursive sequences. Be sure to check our the Evaluating Limits of Recursive Sequences page first for more examples.

## Example 1

Consider the sequence defined recursively such that $a_1 = 2$ and for $n \in \mathbb{N}$, $a_{n+1} = \sqrt{6 + a_n}$. Determine whether this sequence converges or diverges. If it converges, determine to what value.

We suspect that this sequence is both bounded and increasing. If we can show this, then by the monotonic sequence theorem, this sequence will be convergent. We will show that this sequence is both bounded and increasing using the principle of mathematical induction.

First let $P(n)$ be the statement that this sequence is increasing, that is $a_n ≤ a_{n+1}$ for $n \in \mathbb{N}$. $P(1)$ says that $2 ≤ \sqrt{6 + 2} = \sqrt{8}$ which is true. Now suppose that for some $k \in \mathbb{N}$ that $P(k)$ is true, that is $a_k ≤ a_{k+1}$. We want to show that also $P(k+1)$ is true, that is $a_{k+1} ≤ a_{k+2}$. We have that:

(1)
\begin{align} a_{k+2} ≥ \sqrt{6 + a_{k+1}} \overset{IH} ≥ \sqrt{6 + a_k} = a_{k+1} \end{align}

Therefore $P(k+1)$ is true. By the principle of mathematical induction, $P(n)$ is true for all $n \in \mathbb{N}$, or in other words, the sequence $\{ a_n \}$ is increasing.

We now need to show this sequence is bounded. We suspect that this sequence is bounded above by $4$ since $a_1 = 2 ≤ 4$, and $a_2 = \sqrt{8} ≤ 4$, and $a_3 = \sqrt{6 + \sqrt{8}} ≤ 4$. We will now formally prove this, one again by mathematical induction.

Let $S(n)$ be the statement that $a_n ≤ 4$ for $n \in \mathbb{N}$. $S(1)$ says that $2 ≤ 4$ which is true. Now suppose that for some $k \in \mathbb{N}$, that $S(k)$ is true, that is $a_k ≤ 4$. We want to show that also $S(k+1)$ is true, that is $a_{k+1} ≤ 4$. We have that:

(2)
\begin{align} a_{k+1} = \sqrt{6 + a_k} ≤ \sqrt{6 + 4} = \sqrt{10} < 4 \end{align}

Therefore $S(k+1)$ is true. By the principle of mathematical induction, $S(n)$ is true for all $n \in \mathbb{N}$, or in other words, the sequence $\{ a_n \}$ is bounded above by $4$.

So by the monotonic sequence theorem, $\{a_n \}$ is convergent. So we have that $\lim_{n \to \infty} a_n = L$, and so since $\sqrt{6}$ is a continuous function we have that:

(3)
\begin{align} \quad L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{6 + a_n} = \sqrt{6 + \lim_{n \to \infty} a_n} = \sqrt{6 + L} \end{align}

So $L = \sqrt{6 + L}$ which is equivalent to $L^2 - L - 6 = (L -3)(L + 2) = 0$. Therefore $\lim_{n \to \infty} a_n = L = 3$ since $L \neq -2$ (since $L > 0$ since this sequence is positive).