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Lang conjectured that for an irrational algebraic number $\alpha$ and $\epsilon > 0$, there exist only finitely may rationals $p/q$ such that $$ \left| \alpha - \frac{p}{q} \right| <\frac{1}{q^2(\ln q)^{1+\epsilon}} \ .$$ Presumably the heuristic here rests on the observation that the set of all reals that violate this condition has measure zero (which follows because a certain sum of characteristic functions has a finite integral). But this applies equally to $$ \left| \alpha - \frac{p}{q} \right| <\frac{1}{q^2\ln q (\ln \ln q)^{1+\epsilon}}$$ and $$ \left| \alpha - \frac{p}{q} \right| <\frac{1}{q^2\ln q \ln \ln q (\ln \ln \ln q )^{1+\epsilon}},$$ etc.

So, my questions:

1) Do these strengthenings of Roth's theorem/Lang's conjecture appear in the literature? If not, is it because

a) there's no pragmatic point in considering them until someone settles Lang's conjecture or

b) there's some theoretical or known reason actually to doubt their truth.

2) The heuristic above for Lang's conjecture fails if we take $\epsilon=0$, but I don't know where one looks for counter-examples - meaning algebraic numbers (with deg>2) with infinitely many rational approximations that satisfy for some $c$ $$ \left| \alpha - \frac{p}{q} \right| <\frac{c}{q^2\ln q}$$ or if such do exist, then analogous statements with more iterated logarithms in the denominator on the right.

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To be true, nothing is known towards any of your (and Lang's) questions, although there are some false achievements on this way (for example, Tasoev, B. G. On a conjecture of S. Lang. (Russian) Vladikavkaz. Mat. Zh. 4 (2002), no. 2, -- be aware that the proof there is wrong). – Wadim Zudilin Jan 24 '12 at 11:47
Just noticed this long-ago question. I don't recall the exact reference (maybe Lang's little Diophantine Approximation book), but I thought Lang conjectured that if $f(n)$ is any positive real-valued function such that $\sum_{n\ge1} 1/f(n)$ converges, then there are only finitely many solutions to $|\alpha-\frac{p}{q}|<\frac{c}{qf(q)}$. I don't recall his discussing a possible converse as in your question (2). – Joe Silverman Jun 18 at 16:12

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