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 $a_0=3$.


1 Answer 1


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

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .