Evaluating Limits of Recursive Sequences Examples 2

# Evaluating Limits of Recursive Sequences Examples 2

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 $a_1 = 3$ and for $n \in \mathbb{N}$, $a_{n+1} = \sqrt{15 + 2a_n}$. Determine whether this sequence converges or diverges. If it converges, determine to what value.

If we can show that the $\{ a_n \}$ is both increasing and bounded, then the sequence must be convergent.

The first few terms of this sequence are:

(1)
$$a_1 = 3$$
(2)
\begin{align} a_2 = \sqrt{15 + 2(3)} = \sqrt{21} = 4.583... \end{align}
(3)
\begin{align} a_3 = \sqrt{15 + 2\sqrt{21}} \approx 4.915... \end{align}
(4)
\begin{align} a_4 = \sqrt{15 + 2\sqrt{15 + 2\sqrt{21}}} = 4.983... \end{align}

Clearly this sequence is increasing. We will show this by induction. Let $S(n)$ for $n = 1, 2, ...$ be the statement that $a_n < a_{n+1}$. Clearly $S(1)$ is true since $a_1 = 3 < \sqrt{15 + 2(3)} = \sqrt{21} = a_2$. Suppose that $S(k)$ is true for some $k > 1$, that is, assume that $a_k < a_{k+1}$. We want to then show that $S(k+1)$ is true, that is $a_{k+1} < a_{k+2}$. We have that:

(5)
\begin{align} \quad a_{k+1} = \sqrt{15 + 2a_k} \overset{H} < \sqrt{15 + 2a_{k+1}} = a_{k+2} \end{align}

Therefore $S(k+1)$ is true. Therefore by the principle of mathematical induction we have that $a_n < a_{n+1}$ for each $n = 1, 2, ...$. Thus $\{ a_n \}$ is an increasing sequence.

We will now show that the sequence $\{ a_n \}$ is bounded above by $5$. Let $P(n)$ for $n = 1, 2, ...$ be the statement that $a_n < 5$. Clearly $P(1)$ is true since $a_1 = 3 < 5$. Suppose that $P(k)$ is true for some $k > 1$, that is, assume that $a_k < 5$. We want to then show that $P(k+1)$ is true, that is $a_{k+1} < 5$. We have that:

(6)
\begin{align} \quad a_{k+1} = \sqrt{15 + 2a_k} \overset{H} < \sqrt{15 + 2(5)} = \sqrt{25} = 5 \end{align}

Therefore $a_{k+1} < 5$ so $P(k+1)$ is true. Therefore by the principle of mathematical induction we have that $a_n < 5$ for each $n = 1, 2, ...$, that is, $\{ a_n \}$ is a sequenced that is bounded above by $5$.

Therefore since $\{ a_n \}$ is an increasing sequence that is bounded above by $5$ then we have that the sequence $\{ a_n \}$ converges to some limit $L$, and from the continuity of the function $f(x) = \sqrt{15 + 2x}$ we have that:

(7)
\begin{align} \quad L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \sqrt{15 + 2a_n} = \sqrt{15 + 2 \lim_{n \to \infty} a_n} = \sqrt{15 + 2L} \end{align}

Thus $L = \sqrt{15 + 2L}$ so $L^2 = 15 + 2L$ which implies that $L^2 - 2L - 15 = (L - 5)(L + 3) = 0$ so $L = 5$}] or [[$L = -3$. In fact we have that $L \neq -3$ since $\{ a_n \}$ is an increasing sequence starting at $3$, so $L = 5$, that is:

(8)
\begin{align} \quad \lim_{n \to \infty} a_n = 5 \end{align}