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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$2011-01-21
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    In this case you need $a_0$ and $a_1$ nonnegative and at least one of them positive.2011-01-21
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    @Sivaram. $a_{0}$ and $a_{1}$ are surely positive.2011-01-21
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    Are $a_0$ and $a_1$ real?2011-01-21
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    @Peter. Positive and Negative are only defined for reals. So are greater than and less than.2011-01-21
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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).2011-01-21
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    @peter I had never seen lexicographic ordering, how do you order things like 2+i vs 3.2011-01-21
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    @picakhu, $a + bi < c + di$ iff $a < c$ or ($a = c$ and $b < d$).2011-01-21
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    @peter, thanks, that is a total ordering system, is it useful for anything outside of a curiosity?2011-01-21
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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.2011-01-21

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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 $b_n\to 0$.