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

Question

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.

Answer 1

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.

Answer 2

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 $a_{n + 1} - a_n \le 0$.

Answer 3

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 $\lim\limits_{x\to+\infty}f(x)=+\infty$, 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 $\ell=2$.

Answer 4

Here is an alternative elementary way.

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

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

Answer 5

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