I'm trying to prove the convergence of the following sequence: $x_{n+1}=0.5\left(x_n+\dfrac{2}{x_n}\right)$, for $n\geq 1$ and $x_1=2$ using the Monotone Convergence Theorem for sequences (i.e, a sequence is convergent if it is monotone and bounded). I evaluated some values of the sequence to get a handle on the bounds. I got $x_1=2, x_2=1.5, x_3=1.41\bar{6}, x_4=1.414215, x_5=1.4142135$, so it looks like the lower bound is going to be $\sqrt{2}$.
I tried to use induction to prove that $\{x_n\}$ is bounded as follows:
Base Case: $n=1$
Since $x_1=2, \sqrt{2}\leq x_1\leq2$.
The Inductive Step: $n\implies n+1$
Assuming $\sqrt{2}\leq x_n \leq 2$, we want to prove that $\sqrt{2}\leq x_{n+1} \leq 2$.
At this step in the process, I am having trouble finding where $x_{n+1}$ could be bounded.
$x_{n+1}=\dfrac{x_n}{2}+\dfrac{1}{x_n}\\ \leq??$,
but I'm having trouble finding a good bound.
Next, I'd like to show that the sequence is monotonic.
I'm not sure how to start this part of the proof. Perhaps define $f(x_n)=x_{n+1}?$
Once I know that the sequence is both monotone and bounded, I can conclude that the sequence is in fact convergent.
Any hints for the above would be appreciated. Thank you.
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$\begingroup$We are given $$ x_{n+1}=\frac12\left(x_n+\frac2{x_n}\right)\tag{1} $$ Note that if $x\gt0$, then $$ \begin{align} \frac12\left(x+\frac2x\right) &=\frac1{2x}\left(x-\sqrt2\right)^2+\sqrt2\\ &\ge\sqrt2\tag{2} \end{align} $$ Thus, for $n\gt0$, $x_n\ge\sqrt2$. Furthermore, $$ \begin{align} x_{n+1}-x_n &=\frac{2-x_n^2}{2x^n}\\ &\le0\tag{3} \end{align} $$ Thus, $x_n$ is non-increasing and bounded below.
$\endgroup$ $\begingroup$Hint: For the lower bound, apply AM-GM, or prove it directly using algebra.
Hint: For the upper bound, consider the shape of the graph $y = \frac {x}{2} + \frac{1}{x} $. What is the maximum in the domain $[\sqrt{2}, 2]$?
However, you have not shown monotone at all, but just bounded. In these kind of cases, usually you do monotone and bounded at the same time.
Note: If you stare at your sequence, you might realize that $ \sqrt{2} \leq x_{n+1} \leq x_n$. This is somewhat easier to prove.
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