# Does strictly increasing continued-fraction-sequence imply transcendency?

Suppose $a$ is a real number with continued fraction $[0,c_1,c_2,\cdots]$. Suppose, the continued-fraction-sequence is strictly increasing ($c_1<c_2<\cdots$). Can we conclude that $a$ is transcendental ? If not, can we construct a concrete counter-example ?

It is clear that the minimal polynomial of $a$ (if existent) must have degree larger than $2$ because the continued fraction neither terminates nor gets eventually periodic.

• Is there any reason to conclude that it would? – Yuriy S Jan 10 '17 at 0:37
• I haven't at all thought this through, but maybe you could adapt Liouville's argument: there is an upper bound on how well an algebraic number can be approximated by rationals, but your numbers can be much more closely approximated than that limit, so they must be transcendental. – MJD Jan 10 '17 at 0:47
• If the sequence increases strong enough, we have a Liouville-number. On the other hand, unbounded sequences are probably not enough - it is conjectured that the algebraic numbers with minimal degree larger than $2$ have unbounded continued-fraction-sequences. I admit, I do not have a convincing reason for the implication. – Peter Jan 10 '17 at 0:48
• suggest cambridge.org/us/academic/subjects/mathematics/number-theory/… however, most things that you just make up will be unknown. – Will Jagy Jan 10 '17 at 1:44