A recursive sequence is defined by...

Question

A sequence is defined recursively by $a_1=1$ and $a_{n+1} = 1 + \frac{1}{1+a_{n}}$.

Find the first eight terms of the sequence $a_n$. What do you notice about the odd terms and the even terms? By considering the odd and even terms separately, show that $a_n$ is convergent and deduce that its limit is $\sqrt{2}$.

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EDIT: Ok, so I was being an idiot and forgot how to basic math for a little. The first 8 terms are as follows:

EVENS:

a2 = 1.5
a4 = 1.46666...
a6 = 1.415731...
a8 = 1.41426...

ODDS:

a1 = 1
a3 = 1.4
a5 = 1.4054...
a7 = 1.41395...

From these I see that the even terms are decreasing while the odd terms are increasing. How can I use these to prove that $\{a_n\}$ converges? Does it have something to do with alternating series or something similar?

Answer 1

Hint: $\sqrt{2}$ is a piece of cheese, $a_{2n}$ is a slice of bread, $a_{2n+1}$ is another slice of bread. What would you do next?

Answer 2

Hint: Applying the recurrence relation twice, we find that $$ a_{n+2} = 1 + \frac{1}{1+a_{n+1}} = 1 + \frac{1}{1+\left(1 + \frac{1}{1+a_{n}}\right)} = \\ 1 + \frac{1}{2 + \frac{1}{1+a_{n}}}=\\ 1 + \frac{1+a_n}{2 + 2a_n + 1} =\\ 1 + \frac{1+a_n}{3 + 2a_n} = \\ \frac{3 + 2a_n + 1+a_n}{3 + 2a_n} = \\ \frac{4 + 3a_n}{3 + 2a_n} $$ This recurrence gives us both the even and odd sequences. We need to show one of the following:

• The evens decrease to $\sqrt 2$ and the odds increase to $\sqrt 2$
• The evens stay above $\sqrt 2$, the odds stay below $\sqrt 2$, and the difference between the evens and odds converges to $0$.

Answer 3

We have $a_{n+1} = 1 + \frac{1}{1+a_{n}} = \frac{2+a_n}{1+a_{n}} $. Therefore, $a_n > 1$ for all $n$.

Also,

$\begin{array}\\ a_{n+1}-\sqrt{2} &= \dfrac{2+a_n}{1+a_{n}}-\sqrt{2}\\ &= \dfrac{2+a_n-\sqrt{2}(1+a_n)}{1+a_{n}}\\ &= \dfrac{2-\sqrt{2}-a_n(\sqrt{2}-1)}{1+a_{n}}\\ &= \dfrac{(\sqrt{2}-1)(\sqrt{2}-a_n)}{1+a_{n}}\\ \end{array} $

so $\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n} =\dfrac{\sqrt{2}-1}{1+a_{n}} $.

Therefore (1)$a_n-\sqrt{2}$ alternates in sign and (2)$\big|\dfrac{a_{n+1}-\sqrt{2}}{\sqrt{2}-a_n}\big| <\sqrt{2}-1 $. This implies that $a_n-\sqrt{2} \to 0$ so that $\lim_{n \to \infty} a_n =\sqrt{2} $.