Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 2 / b

Rewrite the recurrence relation asSince $a_0=1>0$ and $c>0$, a positive $a_n$ gives a positive $a_{n+1}$. Mathematical induction therefore proves $\boxed{a_n>0\text{ for all }n\ge 0}$, and also shows that every division in the recurrence relation is defined.

For every $n\ge 0$, direct algebra givesThus $\boxed{a_n^2\ge c\text{ for every }n\ge 1}$. This identity is the key estimate in Newton iteration for a square root.

For $n\ge 1$, the preceding parts give a positive denominator and a nonpositive numerator inHence $\boxed{a_{n+1}\le a_n\text{ for every }n\ge 1}$. The restriction $n\ge 1$ matters: if $c>1$, then $a_1=(1+c)/2>a_0$.

The tail $(a_n{)}_{n\ge 1}$ is a monotone sequence bounded below by zero. The bounded monotone sequence theorem therefore gives $a_n\to L\ge 0$. Taking the limit of a sequence in $a_n^2\ge c$ yields $L^2\ge c>0$, so $L>0$. The reciprocal rule from part (a) permits passage to the limit of a sequence in the recurrence relation:Multiplication by $2L$ givesThis proves existence of the positive square root without assuming that existence when constructing the sequence.