# Accelerate convergence of a rational sequence converging to a real number

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?

## 1 Answer

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|

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|

• 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? – The Turtle Heremit Jan 15 at 17:28
• @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. – N. S. Jan 15 at 17:37