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Let $(a_n)_{n \in \Bbb N}$ be a sequence in $\Bbb R$ such that $a_0=3$ and $a_{n+1}=\frac{a_n}{2}+\frac{2}{a_n}$ for $n \geq 0$.

Prove that $a_{n+1} \leq a_n$ for all $n \in \Bbb N$.

First I proved that $a_n \geq 2$ for all $n \in \Bbb N$. And I can see that $a_{n+1} \leq a_n$ is true, but I don't know how to do this in a correct mathematical proof.

And if we have proven $a_{n+1} \leq a_n$ than we know it converges since we also know $a_n \geq 2$, and that $a_0=3$.

Fruitdruif
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    Observe that for $x\ge 2$ we have that $2/x+x/2\le x\iff 2/x\le x/2$. Now call $x=a_n$ –  May 24 '17 at 07:35
  • Closely related: https://math.stackexchange.com/questions/1786553/proving-a-sequence-defined-by-a-recurrence-relation-converges. – Martin R May 24 '17 at 08:06

1 Answers1

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Its easy.
As $$a_n \ge2 \implies \frac{a_n}{2}\ge 1$$ $$\frac{2}{a_n}\le1 \le \frac{a_n}{2}$$ $$\frac{2}{a_n}\le \frac{a_n}{2}$$ $$\frac{2}{a_n}+\frac{a_n}{2}\le \frac{a_n}{2}+\frac{a_n}{2}$$

$$a_{n+1}\le a_n$$

maverick
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