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First some comments on the previous comments: Let $\theta=[0,a_0,a_1,\ldots]$ have convergents $p_i/q_i$. Then the connection between the $a_i$ and how well $p_i/q_i$ approximates $\theta$ is given by the following inequality:

$$\frac{1}{q_k^2(a_{k+1}+2)} < |a-p_k/q_k| \le \frac{1}{q_k^2a_{k+1}}.$$

This is taken from Khintchin's book on continued fractions, Page 36 (with a typo corrected.) This is why Roland's statement is true: For example you can make $\theta:=[0,a_0,a_1,\ldots]$ close to the number $2/3=[0,1,2]$, by taking for $\theta$ something like $[0,1,2,10^{100},\ldots]$.

As for Roth's theorem, if for some $\epsilon>0$ and for infinitely many $k$ it holds that $a_{k+1}\ge q_k^{\epsilon}$, then by the above double inequality, Roth's theorem is violated, and $\theta$ can't be algebraic. But consider $[0,1,2,3,\ldots]$. Here the $q_k$ are bounded below by the Fibonacci sequence and the $a_k$ grow linearly, so there is no problem with Roth's Theorem. For all I can see (SEE THE COMMENTS BELOW!!), $[0,1,2,3,\ldots]$ might be an algebraic number, although that might be ruled out by some of Lang's conjectural strengthenings of Roth.
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First some comments on the previous comments: Let $\theta=[0,a_0,a_1,\ldots]$ have convergents $p_i/q_i$. Then the connection between the $a_i$ and how well $p_i/q_i$ approximates $\theta$ is given by the following inequality:

$$\frac{1}{q_k^2(a_{k+1}+2)} < |a-p_k/q_k| \le \frac{1}{q_k^2a_{k+1}}.$$

This is taken from Khintchin's book on continued fractions, Page 36 (with a typo corrected.) This is why Roland's statement is true: For example you can make $\theta:=[0,a_0,a_1,\ldots]$ close to the number $2/3=[0,1,2]$, by taking for $\theta$ something like $[0,1,2,10^{100},\ldots]$.

As for Roth's theorem, if for some $\epsilon>0$ and for infinitely many $k$ it holds that $a_{k+1}\ge q_k^{\epsilon}$, then by the above double inequality, Roth's theorem is violated, and $\theta$ can't be algebraic. But consider $[0,1,2,3,\ldots]$. Here the $q_k$ are bounded below by the Fibonacci sequence and the $a_k$ grow linearly, so there is no problem with Roth's Theorem. For all I knowcan see, $[0,1,2,3,\ldots]$ is might be an algebraic number! , although that might be ruled out by some of Lang's conjectural strengthenings of Roth.
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First some comments on the previous comments: Let $\theta=[0,a_0,a_1,\ldots]$ have convergents $p_i/q_i$. Then the connection between the $a_i$ and how well $p_i/q_i$ approximates $\theta$ is given by the following inequality:

$$\frac{1}{q_k^2(a_{k+1}+2)} < |a-p_k/q_k| \le \frac{1}{q_k^2a_{k+1}}.$$

This is taken from Khintchin's book on continued fractions, Page 36 (with a typo corrected.) This is why Roland's statement is true: For example you can make $\theta:=[0,a_0,a_1,\ldots]$ close to the number $2/3=[0,1,2]$, by taking for $\theta$ something like $[0,1,2,10^{100},\ldots]$.

As for Roth's theorem, if for some $\epsilon>0$ and for infinitely many $k$ it holds that $a_{k+1}\ge q_k^{\epsilon}$, then by the above double inequality, Roth's theorem is violated, and $\theta$ can't be algebraic. But consider $[0,1,2,3,\ldots]$. Here the $q_k$ are bounded below by the Fibonacci sequence and the $a_k$ grow linearly, so there is no problem with Roth's Theorem. For all I know, $[0,1,2,3,\ldots]$ is an algebraic number!