Limit of a recursive sequence $a_{n+1}=\frac{1}{2}\left(a_n + \frac{5}{a_n}\right)$

Question

A sequence is given by $a_{n+1}=\dfrac{1}{2}\left(a_n + \dfrac{5}{a_n}\right)$ for each integer $n \geq 0$. Does the limit of $a_n$ as $n$ goes to infinity equal $\sqrt{5}$ for $a_0 =2$? What if $a_0=3$?

Answer 1

Fix $k>0$ and $t>0$. I shall prove that the sequence $\left\{a_n\right\}_{n\in\mathbb{Z}_{\geq 0}}$ defined by $a_0=t$ and $$a_{n+1}=\frac{1}{2}\,\left(a_n+\frac{k^2}{a_n}\right)\text{ for every }n=0,1,2,3,\ldots\tag{*}$$ satisfies $$\lim_{n\to\infty}\,a_n=k\,.$$ This problem is a particular case where $k:=\sqrt{5}$ and $t\in\{2,3\}$.

First of all, it is clear that $a_n>0$ for all $n\in\mathbb{Z}_{\geq 0}$. By the AM-GM Inequality, $$a_n=\frac{1}{2}\,\left(a_{n-1}+\frac{k^2}{a_{n-1}}\right)\geq \sqrt{a_{n-1}\left(\frac{k^2}{a_{n-1}}\right)}= k$$ for every $n=1,2,3,\ldots$. Note from (*) that $$a_{n+1}-a_n=\frac{1}{2}\left(\frac{k^2-a_n^2}{a_n}\right)\leq 0\text{ for }n=1,2,3,\ldots\,.$$ That is, the subsequence $\left\{a_n\right\}_{n\in\mathbb{Z}_{>0}}$ is nonincreasing and is bounded from below by $k$. Hence, $\lim\limits_{n\to\infty}\,a_n$ exists (and equals $\inf\big\{a_n\,\big|\,n=1,2,3,\ldots\big\}$). If $L$ is this limit, then $L\geq k>0$. By (*), we have $$L=\frac{1}{2}\,\left(L+\frac{k^2}{L}\right)\text{ or }L=k\,.$$

Answer 2

Generally, assume that $a_0>0$. Then the limit of $a_n$ as $n$ goes to infinity equals $\sqrt5$.

First, $a_1 = \frac{1}{2} ( a_0 + {5 \over a_0}) \geq \sqrt 5$, by using inequality of arithmetic and geometric means.

And by using mathematical induction, we know $a_n \geq \sqrt 5$ for all $n\in\Bbb{N}$. It means that $a_n - \sqrt 5 \geq 0$.

Secondly, $$ \begin{align} a_{n+1} - \sqrt{5} & = \frac{1}{2} (a_n + \frac{5}{a_n}) - \sqrt 5 = \frac{1}{2} (a_n - 2\sqrt{5} + \frac{5}{a_n}) \\ & =\frac{1}{2a_n}({a_n}^2 -2\sqrt5 a_n + 5) = \frac{1}{2a_n} {(a_n - \sqrt 5)}^2 \\ & = \frac{1}{2}(a_n - \sqrt 5)(1- \frac{\sqrt 5}{a_n})\\ & \leq \frac{1}{2}(a_n -\sqrt 5). \end{align} $$

Thus inductively, $ 0 \leq a_{n+1} - \sqrt 5 \leq \frac{1}{2^n} (a_1 - \sqrt 5)$, and we get $\lim\limits_{n \rightarrow \infty} {a_n - \sqrt 5} = 0$.

Answer 3

If the sequence converges to some $a$ from some $a_0$, then

$$a=\frac12\left(a+\dfrac5a\right),$$ or $$a^2=5.$$

We will now show that the sequence converges from any $a_0\ne0$. Clearly all $a_n$ have the same sign, so it suffices to consider $a_0>0$.

We evaluate the "residue"

$$a_{n+1}-\sqrt 5=\frac12\left(a_n+\dfrac5{a_n}\right)-\sqrt5=\frac{(a_n-\sqrt5)^2}{2a_n}.$$

It turns out that we can get rid of the denominator by

$$\frac{a_{n+1}-\sqrt5}{a_{n+1}+\sqrt5}=\left(\frac{a_n-\sqrt5}{a_n+\sqrt5}\right)^2$$ and we can conclude convergence to $\sqrt5$ (the RHS tends to zero) provided

$$\left|\frac{a_0-\sqrt5}{a_0+\sqrt5}\right|<1.$$

And this holds for all positive $a_0$.

Answer 4

You may proceed as follows using the $\color{blue}{MVT}$ with respect to the function $f(x) = \frac{1}{2}\left( x + \frac{5}{x} \right)$.

Here some useful facts:

• $\color{blue}{(1)}$: $f(\sqrt{5}) = \sqrt{5}$.
• $\color{blue}{(2)}$: For any $x >0$ you get via AM-GM: $f(x) \geq \sqrt{x \cdot \frac{5}{x}} = \sqrt{5}$
• $\color{blue}{(3)}$: $f'(x) = \frac{1}{2}\left( 1- \frac{5}{x^2} \right) \Rightarrow 0 < f'(x) < \frac{1}{2}$ for $x > \sqrt{5}$

So, whatever starting value $a_0 > 0$ you choose for your iteration, because of $\color{blue}{(2)}$ you have $a_1 \geq \sqrt{5}$. So, for any $a_n > \sqrt{5}$ $(n\geq 1)$ you get

$$|a_{n+1} - \sqrt{5}| \stackrel{\color{blue}{(1)}}{=} |f(a_n) - f(\sqrt{5})| \stackrel{\color{blue}{MVT}}{=} |f'(\xi_n)||a_n - \sqrt{5}| \stackrel{\color{blue}{(3)}}{\leq} \frac{1}{2}|a_n - \sqrt{5}|$$

So, for $\color{blue}{any}$ starting value $\color{blue}{a_0 > 0}$ you have convergence to $\sqrt{5}$ because $$|a_{n+1} - \sqrt{5}| \leq \frac{1}{2^n}|a_1 - \sqrt{5}| \Rightarrow \color{blue}{\boxed{\lim_{n\to\infty}a_n = \sqrt{5}}}$$

Answer 5

Disclaimer: I'm in a hurry so I'll just show, if the base value is acceptable, that the sequence converges but I don't think its hard to find the limit using subsequences.

$\textbf{Claim 1}: \forall y>0, y+25/y\geq 10$.

$\textit{Proof:} 0\geq (y-5)^2=y^{2}-10y+25\Rightarrow y^{2}+25\geq 10y\Rightarrow y+y/25\geq 10$.

$\textbf{Theorem}: \text{The sequence} \hspace{.5pc} (x_{n}), x_{n}=\displaystyle \frac{1}{2}\left(x_{n}+\frac{5}{x_{n}} \right)$ converges.

$\textit{Proof:} \forall x>0, \displaystyle \left(x+\frac{5}{x}\right)^2=x^{2}+\frac{25}{x^{2}}+10\underset{x^{2}>0\wedge\text{Claim 1}}{\geq}10+10=20\Rightarrow x_{n}^{2}=\frac{1}{4}\left(x_{n}+\frac{5}{x_{n}}\right)^{2}\geq \frac{1}{4}\cdot 20=5\Rightarrow x_{n}^{2}\geq 5\forall n\Rightarrow x_{n}-x_{n+1}=x_{n}-\frac{1}{2}\left(x_{2}+\frac{5}{x_{n}}\right)=\frac{1}{2}x_{n}-\frac{5}{x_{n}}=\frac{x_{n}^{2}-5}{2x_{n}}\underset{x_{n}^{2}\geq 5}{\geq}0\Rightarrow (x_{n})$ is decreasing, bounded above by $x_{1}$ (if base value is appropriate) and bounded below by $\sqrt{5}$ (why?). Then $(x_{n})$ converges [1].

What this means is that we can find a sequence of rationals that approximate square roots. Can you generalize this idea to any $\sqrt{c}, c>0$?

I didn't check out the source but just google monotone convergence theorem or trust wiki :)

[1] https://en.wikipedia.org/wiki/Monotone_convergence_theorem

[2] Understanding Analysis; Stephen Abbott, 2nd ed.