# example of contractive sequence

Define the sequence^{} ${a}_{1},{a}_{2},{a}_{3},\mathrm{\dots}$ by

${a}_{1}:=1,{a}_{n+1}:=\sqrt{5-2{a}_{n}}\mathit{\hspace{1em}}(n=1,2,3,\mathrm{\dots}).$ | (1) |

We see by induction^{} that the radicand in (1) cannot become
negative; 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 for an ${a}_{n}$, it implies that $$, i.e. $$.

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^{}

${a}_{n}-{a}_{n+1}$ | $={\displaystyle \frac{(\sqrt{5-2{a}_{n-1}}-\sqrt{5-2{a}_{n}})(\sqrt{5-2{a}_{n-1}}+\sqrt{5-2{a}_{n}})}{\sqrt{5-2{a}_{n-1}}+\sqrt{5-2{a}_{n}}}}$ | ||

$={\displaystyle \frac{-2({a}_{n-1}-{a}_{n})}{\sqrt{5-2{a}_{n-1}}+\sqrt{5-2{a}_{n}}}}$ |

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

$$|{a}_{n}-{a}_{n+1}|=\frac{2|{a}_{n-1}-{a}_{n}|}{\sqrt{5-2{a}_{n-1}}+\sqrt{5-2{a}_{n}}}\leqq \frac{2|{a}_{n-1}-{a}_{n}|}{\sqrt{5-2\sqrt{3}}+\sqrt{5-2\sqrt{3}}}=\frac{|{a}_{n-1}-{a}_{n}|}{\sqrt{5-2\sqrt{3}}}$$ |

Since $$, our sequence (1) is
contractive, consequently Cauchy. Therefore it converges^{} to
a limit $A$.

We have

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

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

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