Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 12F / ii

Let ${\xi }_{n+1}=a_{n+1}+1/(a_{n+2}+\cdots )$ be the complete quotient. Iterating the fractional-linear formula for a continued fraction givesThe supplied determinant identity and ${\xi }_{n+1}>a_{n+1}$ therefore implyThe positive partial quotients ensure increasing denominators tending to infinity; the same determinant identity makes the even and odd convergents nested, giving their common limit $x$. The displayed formula follows by passing to the limit in the finite-tail identity.

An explicit choice is $a_n=10^{n!}$ for every $n\ge 1$. The recurrence gives $q_n\le {\prod }_{j=1}^n(a_j+1)\le 2^{n}10^{{\sum }_{j=1}^{n}j!}$. For $n\ge 2$, ${\sum }_{j=1}^{n}j!\le 2n!$, so ${\log }_{10}{q}_n\le 3n!$. Consequently $a_{n+1}=10^{(n+1)!}\ge q_n^{(n+1)/3}$ andFirst $x$ is irrational: if $x=P/Q$, the nonzero difference from a convergent is at least $1/(Q{q}_n)$, contradicting the error bound once $q_{n+1}>Q$. If $x$ were algebraic of any fixed degree $d$, the last displayed upper bound would eventually be smaller than $c/q_n^d$ for the positive constant in part (i), since $q_n\to \infty$. This contradicts the Liouville approximation theorem. Thus this explicit continued fraction is transcendental.