Ytukyg

Prove the convergence of the sequence $a_{1} = 4$, $a_{n + 1} = frac{a_{n}}{2} + frac{2}{a_{n}}$, $n = 1, 2, ldots$

I'm pretty sure the way to do it is to show $a_{n} > 2$ for $n = 2, ldots$ and then maybe use the Monotone Convergence Theorem to show it converges to $2$, but I think this also might be wrong. Can someone please help me with this problem? I don't know how to prove a bound for it.

Your idea is a good approach. The fact that $a_{n} ge 2$ holds for all $n$ follows simply from the fact that $a_n > 0$ for all $n$, and the minimum of $t + 1/t$ on the interval $(0, infty)$ is $2$.

As far as monotonicity, note that

$$a_{n + 1} - a_n = frac{2}{a_n} - frac{a_n}{2} = t - frac 1 t$$

where $t = 2 / a_n in (0, 1]$ from above. The maximum of $t - 1/t$ on this interval is zero, so $a_{n + 1} - a_n le 0$.

The recurrence relation depends on the function $f(x)=dfrac x2+dfrac2x$. Its derivative
$$f'(x)=frac12-frac 2{x^2}=frac{x^2-4}{2x^2}$$
shows it is increasing on $[2,+infty)$. As $f(2)=2$ and $limlimits_{xto+infty}f(x)=+infty$, it maps the interval $I=[2,+infty)$ into itself, so the sequence is bounded from below.

Furthermore, you easily check $f(x)<x$ on $(2,+infty)$, so there results the sequence is decreasing. By the monotone convergence theorem, it converges to a (non-negative) fixed point of the function. The single such point is $ell=2$.

Making it more general, rewrite
$$a_{n + 1} = frac{a_{n}}{2} + frac{k}{a_{n}}$$ as
$$a_{n + 1} =a_n- frac{a_{n}}{2} + frac{k}{a_{n}}=a_n-frac{a_n^2-2k}{2a_n}$$ and recognize the formula of Newton iterates for finding the zero of $f(x)=x^2-2k$.

This is the the so-called Babylonian method.

Here is an alternative elementary way.

Note that for $x,ygeqslant 0$, by the AM-GM inequality (or simply noticing that $(sqrt{x}-sqrt{y})^2geqslant 0$),
$$
x+ygeqslant 2sqrt{xy}.
$$
One can easily show by induction that $a_ngeqslant 0$ for all $n$. So for all $n>1$
$$
a_{n+1}geqslant 2sqrt{frac{a_n}{2}cdotfrac{2}{a_n}}=2,
$$
and thus $a_ngeqslant 2$ for all $ngeqslant 1$.

For monotonicity, simply note that $a_ngeq 2$ implies that
$$
frac{2}{a_n}leqslant 1,quad -frac{a_n}{2}leqslant -1,
$$
and thus
$$
a_{n+1}-a_n=frac{2}{a_n}-frac{a_n}{2}leqslant 1+(-1)=0.
$$

Note, that $color{blue}{a= 2}$ is a $color{blue}{mbox{fixpoint}}$ of the iteration as
$$2 = frac{a}{2} + frac{2}{a} = 1+1$$

AM-GM shows that for all members of the sequence we have
$$frac{a_n}{2} + frac{2}{a_n} stackrel{AM-GM}{geq} 2$$

Now, consider
$$0 leq color{blue}{a_{n+1}-2} = frac{a_n}{2} + frac{2}{a_n} - 2 = frac{a_n -2}{2} - frac{a_n - 2}{a_n} = left( frac{1}{2} - frac{1}{a_n} right)(a_n - 2) color{blue}{stackrel{a_n geq 2}{leq} frac{1}{2} (a_n - 2)}$$
It follows
$$0 leq a_{n+1}-2 leq left( frac{1}{2}right)^n(a_1 - 2)stackrel{n to infty}{longrightarrow} 0 Rightarrow color{blue}{(a_n) mbox{ is convergent and } lim_{n to infty}a_n = 2}$$