By standard density argument we know that the rationals are dense in the reals, so for every real number I can extract a sequence converging to it. So we can say that $$ \lim_{n \rightarrow \infty} q_n - r = 0$$

My question is, is it always possible to extract a subsequence from my generic sequence $(q_n)$, such that the convergence of the subsequence to the same limit $r$ is faster then the original? (I mean, the convergence of $(q_{n_j})$ to $r$ is asymptotically equivalent to $\frac{1}{n^2}$ for example?


The answer is yes.

Pick any rate of convergence $r_n$ you want. Then, for each $n$ there exists some $M_n$ such that for all $m >M_n$ we have $$|a_n-l|<r_n$$

Pick inductively $$k_n > \max \{ k_1,k_2,..., k_{n-1}, M_n \}$$

Then $a_{k_n}$ is a subsequence and since $k_n >M_n$ we have $$|a_{k_n} -l| <r_n$$

  • $\begingroup$ So we are basically choosing the natural number $M_n$ so that we make $\epsilon$ small enough (what you called $r_n$) to convergence, at every step, at the rate I want? $\endgroup$ – The Turtle Heremit Jan 15 at 17:28
  • 1
    $\begingroup$ @TheTurtleHermit For each $n$ you set $\epsilon$ to be the rate of convergence at that $n$ and pick one element which satisfies that inequality, yes. $\endgroup$ – N. S. Jan 15 at 17:37

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