Series definied by recurrence relation

Question

Define a sequence recursively by $$\large{a_1 = \sqrt{2},a_{n+1} = \sqrt{2 + a_n}}$$

(a) By induction or otherwise, show that the sequence is increasing and bounded above by 3. Apply the monotonic sequence theorem to show that limit as $\large{n \to \infty}$ exists.

(b) What is the value of limit as $\large{n \to \infty}$ of $\large{a_n}$? Carefully justify your answer.

I've spent some time analysing this problem. The question asks to prove that it is bounded above by three, but I am almost certain it is bounded above by $2$ as well. Since this is a stronger bound am I correct in thinking I should use this in my solution? I have not completed any problems with a recursively defined series before so I am unsure how to even start this part of the proof.

To show that the series is increasing I know that I need to show that for all $\large{n, a_n+1 > a_n}$.

I've been told that I need to do two induction proofs for part a) of the question. One to prove its bounded and the other to prove it's increasing.

Applying the monotonic sequence is easy because once you know its bounded and increasing then it must converge. It's the first part of the problem I am having trouble with.

Answer 1

That $a_2>a_1$ is clear by squaring both sides. So assume $a_n>a_{n-1}$. Want $a_{n+1}>a_n$. This equivalent to $\sqrt{2+a_n}> \sqrt{2+a_{n-1}}$ and this is true by squaring both sides and using induction hypothesis.

Now for $0<a_n<3$. It is true for $n=1$.

Assume true for $n$. Now $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+3}<3$.

Now for the limit. The limit a exists by monotonicity and boundedness. From $a_{n+1}=\sqrt{2+a_n}$ and upon taking $n\to\infty$ we see that $a=\sqrt{2+a}$ and so $a^2-a-2=0$ and so $a=2$ since $a_1>1$ and the sequence is increasing.

Answer 2

To prove that $(a_n)_n$ is an increasing series, we first observe that \begin{equation*} a_1 = \sqrt{2} < \sqrt{2 + \sqrt{2}} = a_2. \end{equation*} Assume now that $a_n$ > $a_{n-1}$. We then have \begin{equation*} a_{n+1} = \sqrt{2+a_n} > \sqrt{2+a_{n-1}} = a_n, \end{equation*} since $a_n > a_{n-1}$ and $f(x) = \sqrt{x}$ is an increasing function. We have shown that $a_{n+1} > a_n$ for all $n\in \mathbb N$.

For the upper bound, we note that \begin{equation*} a_1 = \sqrt{2} < 2. \end{equation*} We assume that $a_n < 2$ and obtain \begin{equation*} a_{n+1} = \sqrt{2+a_n} < \sqrt{2 + 2} = 2, \end{equation*} since $a_n < 2$ and $f(x) = \sqrt(x)$ is increasing.

We now know that $(a_n)_n$ is increasing and bounded. Hence it has a limit $a$. If we now take $n \rightarrow \infty$ we have \begin{align*} a &= \sqrt{2+a},\\ a^2 &= 2 + a,\\ a &= 2. \end{align*} Note that the condition for the upper bound $a_n < 2$ changes to $a \leq 2$ if we take $n\rightarrow \infty$.