example of contractive sequence

Define the sequence $a_{1},a_{2},a_{3},\ldots$ by

{\displaystyle a_{1}:=1,\qquad a_{n+1}:=\sqrt{5-2a_{n}}\quad(n=1,2,3,\ldots}).

(1)

We see by induction that the radicand in (1) cannot become negative; in fact we justify that

\displaystyle 1\leqq a_{n}\leqq\sqrt{3}

(2)

for every $n$ : It’s clear when $n=1$ . If it is true for an $a_{n}$ , it implies that $1<5-2a_{n}\leqq 3$ , i.e. $1<a_{n+1}\leqq\sqrt{3}$ .

As for the convergence of the sequence, which is not monotonic, one could think to show that it is a Cauchy sequence. Unfortunately, it is almost impossible to directly express and estimate the needed absolute value of $a_{m}-a_{n}$ . Fortunately, the recursive definition (1) allows quite easily to estimate $|a_{n}-a_{n+1}|$ . Then it turns out that it’s a question of a contractive sequence, whence it is by the parent entry (http://planetmath.org/ContractiveSequence) a Cauchy sequence.

We form the difference

	$\displaystyle a_{n}-a_{n+1}$	$\displaystyle=\frac{(\sqrt{5-2a_{n-1}}-\sqrt{5-2a_{n}})(\sqrt{5-2a_{n-1}}+% \sqrt{5-2a_{n}})}{\sqrt{5-2a_{n-1}}+\sqrt{5-2a_{n}}}$
		$\displaystyle=\frac{-2(a_{n-1}-a_{n})}{\sqrt{5-2a_{n-1}}+\sqrt{5-2a_{n}}}$

where $n>1$ . Thus we can estimate its absolute value, by using (2):

|a_{n}-a_{n+1}|=\frac{2|a_{n-1}-a_{n}|}{\sqrt{5-2a_{n-1}}+\sqrt{5-2a_{n}}}% \leqq\frac{2|a_{n-1}-a_{n}|}{\sqrt{5-2\sqrt{3}}+\sqrt{5-2\sqrt{3}}}=\frac{|a_{% n-1}-a_{n}|}{\sqrt{5-2\sqrt{3}}}

Since $\frac{1}{\sqrt{5-2\sqrt{3}}}<1$ , our sequence (1) is contractive, consequently Cauchy. Therefore it converges to a limit $A$ .

We have

A^{2}=\left(\lim_{n\to\infty}a_{n+1}\right)^{2}=\lim_{n\to\infty}a_{n+1}^{2}=% \lim_{n\to\infty}(5-2a_{n})=5-2A.

From the quadratic equation $A^{2}+2A-5=0$ we get the positive root $A=\sqrt{6}-1$ . I.e.,

\displaystyle\lim_{n\to\infty}a_{n}=\sqrt{6}-1.

(3)