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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?

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@Vinod: You need to specify the initial conditions i.e. in this case $a_0$ and $a_1$ –  user17762 Jan 21 '11 at 7:46
    
In this case you need $a_0$ and $a_1$ nonnegative and at least one of them positive. –  Jonas Meyer Jan 21 '11 at 7:48
    
@Sivaram. $a_{0}$ and $a_{1}$ are surely positive. –  Vinod Jan 21 '11 at 7:56
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@picakhu, my point really was that it's polite to state your definitions up front rather than making people deduce them from information contained more than half-way through. But as to greater than and less than, there's a natural total ordering of the complex numbers (lexicographic ordering based on the natural ordering of the reals). –  Peter Taylor Jan 21 '11 at 16:53
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@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. –  Peter Taylor Jan 21 '11 at 18:55
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1 Answer

up vote 5 down vote accepted

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 $b_n\to 0$.

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