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We define $a_1=1$ and $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)$.

I want to prove that $a_n$ converges. But first I’d like to show that $a_n<\sqrt{2}$ for every $n\in\mathbb{N}$.

I try to use induction. For a fixed $n$, I suppose that $a_n<\sqrt{2}$.

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Any hint to prove that $a_{n+1}=\dfrac{1}{2}\left(a_n+\dfrac{2}{a_n}\right)<\sqrt{2}$?

Thanks.

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We have:

$$ 2 a_n a_{n+1} = a_n^2 + 2, $$

from which it follows that:

$$ a_{n+1}^2-2 = (a_{n+1}-a_n)^2 = \left(\frac{1}{a_n}-\frac{a_n}{2}\right)^2 = \left(\frac{a_n^2-2}{2a_n}\right)^2 > 0, $$

so $a_n > \sqrt{2}$ for every $n > 1$. If we set $b_n = a_{n+1}^2-2$, we have $b_1=\frac{1}{4},b_n>0$ and:

$$ b_n = \frac{b_{n-1}^2}{4a_n^2} = \frac{b_{n-1}^2}{4(b_{n-1}+2)} \leq \frac{b_{n-1}^2}{8}, $$

from which it follows that the sequence $\{b_n\}_{n\geq 1}$ decreases very fast towards zero. For instance, $b_1=\frac{1}{4}$ and the last inequality imply, by induction:

$$ b_n \leq \frac{1}{2^{5\cdot 2^{n-1}-3}}.$$

In a general case :

$$a + b – 2\sqrt{ab} = (\sqrt{a} – \sqrt{b})^2 \geq 0,$$

so just put $a = a_n, b=\frac{2}{a_n}$ in your case.

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