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In one of the construction of $\mathbb{R}$ we make each real number an equivalence class of Cauchy sequences in $\mathbb{Q}$. More precisely, two Cauchy sequences $a_n$ and $b_n$ are equivalent iff $|a_n-b_n|\to 0$.

Is there a general method of computing the decimal expansion of the limit of a Cauchy sequence? In other words, given a Cauchy sequence $a_n$ is there a way of determining an equivalent Cauchy sequence $b_n$ such for all $n\in\mathbb{N}$ it is true that $b_{n+1}$ and $b_n$ are identical for the first $n$ digits (making $b_n$ a sequence of decimal truncation of $\lim_{n\to\infty} a_n$). If there is no such algorithm, can we determine these decimal places up to $n$ digits? My problem is that different Cauchy sequences converge at various rates, and determining the nature of their convergence seems to be a case-by-case problem.

If there is no algorithm, how can one go about proving that every real number (as equivalence classes of Cauchy sequences) has a decimal expansion?

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If you have constructive info about the Cauchy-ness of the sequence, i.e. you are able for any given $\epsilon>0$ to actually exhibit $N$ such that $|a_m-a_n|<\epsilon$ for all $n,m>N$, you may have a chance to algorithmically find the decimal (or binary or whatever) expansion of the limit. To find the first $n$ digits, you can start with some $\epsilon<10^{-n}$, determine the corresponding $N$, check how close $a_{N+1}$ is to adjacent $n$-digit numbers and retry with smaller $\epsilon$ if necessary. This is guaranteed to give you the desired $n$ digits after finite time - unless the limit actually happens to be precisely of the form $\frac m{10^n}$.

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For the second part of your question, if you have $a\in\mathbb R$ you can build the sequence $a_n=\lfloor a\cdot10^n\rfloor\cdot10^{-n}$ and then prove that this is a Cauchy sequence.

Now, you are right about different rates of convergence. Given $a_n$ a Cauchy sequence, then $b_n=\lfloor a_n\cdot10^n\rfloor\cdot10^{-n}$ should also be a Cauchy sequence converging to the same point a $a_n$, but if $a_n$ converges slowly, then $b_n$ and $b_{n+1}$ will have few ($\ll n$) digits in common.

Another possibility is having the subseries $\varepsilon_m=10^{-m}$, now, by definition of Cauchy sequence, there should be $N_m$ such as $|a_n-a_{N_m}|<\varepsilon_m$ (for each $n>N_m$), so let's take $b_m=\lfloor a_{N_m}\cdot10^m\rfloor\cdot10^{-m}$ and it should be possible to prove that $b_m$ is a Cauchy sequence and it converges as $a_n$.

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  • $\begingroup$ This is pretty close, but I would specifically like to have $b_n$ represent decimal truncations of $a=\lim_{n\to\infty} a_n$. Also sorry for the really late response. $\endgroup$ – Christian Bueno Sep 27 '14 at 8:49
  • $\begingroup$ @ChristianBueno The second $b_n$ sequence is pretty close to the decimal expansion of $a$, although in some cases the $n$th digit might be off by 1. $\endgroup$ – Carlos Eugenio Thompson Pinzón Sep 30 '14 at 23:07

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