Vtyjkyuk

The simple continued-fraction-expansion for the transcendental number $e$ is known to be unbounded. What about bounded continued fractions ?

Do we know any transcendental number for which it is proven that the simple continued-fraction-expansion is bounded ?

It is conjectured that the simple continued-fraction-expansion of the algebraic numbers with minimal polynomial degree greater than $2$ are unbounded.

If this would be true, every bounded non-periodic infinite simple continued-fraction-expansion would correspond with a transcendental number.

But to my knowledge, it was not proven for a single algebraic number with minimal polynomial degree greater than $2$, that its simple continued-fraction-expansion is unbounded.

Yes, but the transcendentals that this answer describes are quite unnatural.

Fix any noncomputable bounded sequence of positive integers $alpha$, and let $r_alpha$ be the real number whose continued fraction expansion is given by $alpha$. Then - since the continued fraction expansion of a computable real is computable, and every algebraic real is computable - $r_alpha$ is transcendental. Note that the nontrivial part here is proving transcendentality!

There must be lots of them. There are uncountably many sequences consisting of $1$s and $2$s. If you consider them as continued fractions, at most countably many are algebraic, so most of them are transcendental. That doesn't let me pick out a specific one, however, as transcendental.