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May 9, 2016 at 17:33 answer added Robert Israel timeline score: 3
May 9, 2016 at 15:56 history edited Myshkin
+ top level tag (nt.)
May 8, 2016 at 11:54 history edited Siminore CC BY-SA 3.0
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May 8, 2016 at 11:32 comment added Siminore @benblumsmith Do you think it is so easy? Take for instance $\varepsilon_k = 2^{-k}$. Then we want $q_k \lesssim 2^k$. I don't see how a pigeonhole principle can be invoked. How can we bound, a priori, the largeness of the integer $q_k$? In some sense there are too few integers below $2^k$...
May 8, 2016 at 11:29 comment added benblumsmith I think this should be true using some form of the pigeonhole principle, like Dirichlet's approximation theorem.
May 8, 2016 at 11:18 comment added Siminore @joro $\lim_k = \lim_{k \to +\infty}$, the only possible limit as $k \in \mathbb{N}$.
May 8, 2016 at 11:17 history edited Siminore CC BY-SA 3.0
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May 8, 2016 at 11:10 comment added joro What does $\lim_k$ mean?
May 8, 2016 at 10:57 history asked Siminore CC BY-SA 3.0