2
$\begingroup$

Given the positive sequence $a_{n+2} = \sqrt{a_{n+1}}+ \sqrt{a_n}$,

I want to prove these.

1) $|a_{n+2}| > 1 $ for sufficiently large $n \ge N$.

2) Let $b_{n} = |a_{n} - 4|$. Show that $b_{n+2} < (b_{n+1} + b_{n})/3$ for $n \ge N$.

3) Prove that the sequence converges.

How should I proceed? Is there a recurrence formula for $a_{n}$ like a continued fraction?

  • 2
    @picakhu, off the top of my head I can't think of anything, but total orderings in general can be handy in computation when you want canonical representations of things in general.2011-01-21

1 Answers 1

5

Here are some hints.

1) You can show that if $a_{n}$ is ever greater than $1$, then so is $a_{n+1}$. On the other hand, if $a_n<1$, then you can show that $a_{n+1}\gt a_n$, and $a_{n+2}\gt 2a_n$.

2) You can use the triangle inequality and a factorization trick, $\sqrt{x}-\sqrt{4}=\frac{x-4}{\sqrt{x}+\sqrt{4}}$.

3) Show that $b_n\to 0$.