2
$\begingroup$

$a_{n+1} = 1-\frac{1}{a_n + 1}$
$a_1 = \sqrt 2$

I need to prove:
1. $a_n$ is irrational for every $n$.
2. $a_n$ convregres

My ideas:

  1. Induction - as I know that $a_1 $ is irrational, I'm assuming $a_n$ is also irrational, and then: $a_{n+1} = 1-\frac{1}{a_n + 1} \implies a_{n+1} = \frac{a_n}{a_n + 1}$. Now it can be explained that $\frac{a_n}{a_n + 1}$ has got to be irrational.
  2. No good ideas here, I can show that $a_n > 0$ for every $n$ by induction. Then $\frac{a_{n+1}}{a_n} = \frac{1}{a_n + 1} \le 1$. No idea how to continue
  • 3
    To more easily explain why $a_{n+1}$ must be irrational, express $a_n$ in terms of $a_{n+1}$. Then you can see that if $a_{n+1}$ is rational, $a_n$ must be rational. It's not _entirely_ obvious why $a_n$ irrational implies that $\frac{a_n}{a_n+1}$ is rational. – 2017-02-17

2 Answers 2

5
  1. If $a_n$ is irrational, then suppose that $a_{n+1} = \frac{a_n}{a_n + 1}=\frac{p}{q}$ it follows $a_n= \frac{p}{q-p}$.

This gives: if $a_n \notin \mathbb Q$ then $a_{n+1} \notin \mathbb Q$ .

  1. You have shown that $ \frac{a_{n+1}}{a_n} = \frac{1}{a_n + 1} \le 1$, hence $a_{n+1} \le a_n$.

Therefore $(a_n)$ is monotonic.

From

$0

Conclusion: $(a_n)$ is convergent.

  • 0
    Nitpick: what the OP must be wanting is: the sequence's monotonic *descending* and bounded *from below* .+1 – 2017-02-17
1

Show convergence by proving that $a_n$ is bounded below by $0$, as you already did. Then you can show, that the sequence is drecreasing by starting with $a_{n+1}