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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

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.

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    Is there any reason to conclude that it would? – 2017-01-10
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    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. – 2017-01-10
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    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. – 2017-01-10
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    suggest http://www.cambridge.org/us/academic/subjects/mathematics/number-theory/neverending-fractions-introduction-continued-fractions however, most things that you just make up will be unknown. – 2017-01-10

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