Showing that $a_1=2$ and $a_n=\frac{1}{2}(a_{n-1}+\frac{2}{a_{n-1}})$ define a decreasing bounded sequence

Question

I need to show that the sequence given by $$a_1=2$$ $$a_n=\frac{1}{2}\Bigg(a_{n-1}+\frac{2}{a_{n-1}}\Bigg)$$ is monotonically decreasing and bounded. But every time I try, I can't make the bounds tight enough to prove it. How would I solve this?

Answer 1

Here $a_n$ satisfies the quadratic equation $$a_n^2-(2a_{n+1})a_n+2=0$$ This equation has a real root and hence the discriminant $$4(a_{n+1}^2-2) \geq 0$$ That is, $$a_{n+1}^2 \geq 2$$ for $n \geq 1$.

Now,
$a_{n}-a_{n+1}=a_{n}-\frac{1}{2}\Bigg( a_n+\frac{2}{a_n} \Bigg)=\frac{1}{2}\Bigg(\frac{a_n^2-2}{a_n} \Bigg)\geq 0$

for all $n \geq 1$.

Hence $$a_{n+1} \leq a_n, \forall n \geq 2$$

and so by MCT it converges to its glb!

Answer 2

Rate of Convergence

The limit is $\sqrt2$. In fact, the convergence is quite fast, about doubling the number of digits each iteration: $$ a_n-\sqrt2=\frac1{2a_{n-1}}\left(a_{n-1}-\sqrt2\right)^2\tag1 $$

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Nature of Convergence

Note that $(1)$ guarantees that $a_n\ge\sqrt2$ for $n\ge2$. Furthermore, $$ a_n-a_{n-1}=\frac1{2a_{n-1}}\left(2-a_{n-1}^2\right)\tag2 $$ which, because $a_{n-1}\ge\sqrt2$, guarantees that $a_n\le a_{n-1}$ for $n\ge3$.

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Since $a_1=2\gt\sqrt2$, $a_n$ is decreasing and bounded below by $\sqrt2$.

Answer 3

$a_{n+1}-a_n=\frac12\left(a_n+\frac2a_n\right)-a_n=-\frac12\left (a_n-\frac2{a_n}\right) =-\frac12\left(\frac{a_n^2-2}{a_n}\right)\le0$, since $a_n\ge\sqrt2$ (easy to see).

Hence $(a_n)$ is monotonically decreasing.

The limit $a$ satisfies $$a=\frac12\left(a+\frac2a\right)\implies 2a^2=a^2+2\implies a^2=2$$.

Hence $a_n\to\sqrt2$.