Do we know a transcendental number with a proven bounded continued fraction expansion?

Question

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.

Answer 1

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

Here's one for you:

$\begin{align} \kappa &= \sum^\infty_{n=0}10^{-2^{n}} \\ &= 0.\mathbf{1}\mathbf{1}0\mathbf{1}000\mathbf{1}0000000\mathbf{1}000000000000000\mathbf{1}0000000000000000000000000000000\mathbf{1}\ldots \end{align}$

$\kappa$ has been proven to be transcendental. And its canonical continued fraction is:

$\left[0; 9, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 8, 10, 12, 10, 10, 8, 10, 12, 8, 10, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, 12, 10, 8, 12, 10, 8, 10, 10, 12, 10, 8, 12, 10, 10, 8, 10, \ldots\right]$

With the exception of the first two terms which are 0 and 9, respectively, $\kappa$'s continued fraction contains no other terms than 8, 10, or 12.

If we write this as $\left[a_0;a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8,\ldots\right]$, and noting that $\left(\frac{n}{m}\right)$ represents the Jacobi symbol, then we can actually deduce what each term is:

$\forall~n\in\mathbb{Z}_{\geqslant 0},~a_n=\begin{cases} 0 & n=0 \\ 8 & n\in\left\{\frac{8m+\left(\frac{-1}{m-1}\right)+1}{2}~:~m\in\mathbb{Z}^{+}\right\} \\ 9 & n=1 \\ 10 & \text{otherwise} \\ 12 & n\equiv 2\left(\operatorname{mod}8\right)\text{or}~7\left(\operatorname{mod}8\right) \end{cases}$

So there we have it. A transcendental number whose continued fraction has bounded terms. In fact, it is conjectured that all continued fractions that are infinite, bounded in their terms, and not eventually periodic produce transcendental numbers.

Answer 2

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!