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As we know, real numbers are constructed by filling up gaps between rationals by the limits of all Cauchy sequences of rationals.

Q. What are examples of sequence of rationals such that its easy to see that their limit is not a rational and the construction of such sequences is easier, natural or are of geometric flavour than ad-hoc construction?

One may immediately think of $\sqrt{2}$ as one number, and go to construct a sequence for it.

Or one may take $1, 1+ \frac{1}{2!}, 1+\frac{1}{2!} + \frac{1}{3!}, \cdots$ and show that limit of this sequence is not a rational; but the proof of this requires non-natural tricks!

But I am considering reverse way: whether we can produce easily many examples of Cauchy sequences of rationals easily whose limit is not a rational and easy to illustrate to beginners of undergraduate students?

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    $\begingroup$ i mean you can just fill out decimal expansion of irrational number. it won't be periodic, so clearly won't converge to rational number $\endgroup$ – mathworker21 Mar 25 '19 at 7:06
  • $\begingroup$ Or use the convergents of a continued fraction expansion of the irrational, but that might be to advanced too. $\endgroup$ – Henno Brandsma Mar 25 '19 at 7:08
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In general, it is hard to prove that a specific number is irrational, and a lot of irrationality proofs, especially those involving limiting sequences, involve "tricks" like the one you mentioned for $e$. However, there are some "artificial" irrational numbers that are constructed as a limit whose irrationality can be proven via the following not-that-hard-to-believe result:

A real number is rational if and only if its decimal expansion is eventually periodic.

This should not be too hard to convince students of. Now, we can construct numbers that have clearly aperiodic decimal expansions like

$$1.01001000100001000001\dots$$

as limits of sequences like

$$1,1.01,1.01001,1.010010001,\dots$$

whereupon it should be obvious via the previous result that the above number is irrational.

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For each $r\in\mathbb Q^+$ such that $\sqrt r\notin\mathbb Q$, you can consider the sequence $(a_n)_{n\in\mathbb{Z}^+}$ defined by

  • $a_0=1$;
  • $\displaystyle a_{n+1}=\frac12\left(a_n+\frac r{a_n}\right)$.

Then each $a_n$ is rational and $\lim_{n\to\infty}a_n=\sqrt r\notin\mathbb Q$.

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  • $\begingroup$ Do you mean $\lim_{n\rightarrow\infty} a_n = \sqrt r$? $\endgroup$ – o.h. Mar 25 '19 at 7:17
  • $\begingroup$ Yes. I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Mar 25 '19 at 7:19
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If $s\in\mathbb R$, then for all $n\in\mathbb N$, it is clear that $\dfrac{\lfloor ns\rfloor}{n}\in\mathbb Q$, where $\lfloor x\rfloor$ is the integer part of $x$, and $$ \frac{\lfloor ns\rfloor}{n}\to s. $$ One interesting way to obtain convergence to $\sqrt{2}$ with rationals, is by the following recursive sequence $$ a_0=1, \quad a_{n+1}=\frac{1}{2}\left(a_n+\frac{2}{a_n}\right). $$ In particular, the recursive sequence $$ b_0=1, \quad b_{n+1}=\frac{1}{2}\left(a_n+\frac{k}{a_n^{m-1}}\right). $$ converges to $\sqrt[m]{k}$, for all $m,k>0$.

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