example of contractive sequence

Define the sequence  $a_1,a_2,a_3,\mathrm{\dots }$  by

$a_1:=1,a_{n+1}:=\sqrt{5-2a_n}\mathit{\hspace{1em}}(n=1,2,3,\mathrm{\dots }).$(1)

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

$1\leqq a_n\leqq \sqrt{3}$(2)

for every $n$:  It’s clear when $n=1$. If it is true foran $a_n$, it implies that , i.e..

As for the convergence of the sequence, which is not monotonic, one could think to show thatit is a Cauchy sequence. Unfortunately, it is almost impossible to directly express andestimate 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 acontractive sequence, whence it is by the parent entry (http://planetmath.org/ContractiveSequence) aCauchy sequence.

We form the difference

$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, byusing (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 , our sequence (1) iscontractive, consequently Cauchy.  Therefore it converges toa limit $A$.

We have

$$A^2={\left(\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}\right)}^2=\underset{n\to \mathrm{\infty }}{lim}{a}_{n+1}^2=\underset{n\to \mathrm{\infty }}{lim}(5-2a_n)=5-2A.$$

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

$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\sqrt{6}-1.$(3)