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Consider the sequence of partial sums $s_n = \sum_{k=1}^{n} \frac{1}{k^2}$. As a finite sum of rational numbers, the $s_n$ itsself are rational. However, since $\pi$ is transcendental, $\pi^2$ is an irrational number.

We can show that $s_n$ is Cauchy and $s_n \to \pi^2/6$, i.e. we have a cauchy sequence lying in $\mathbb{Q}$, which does not converge in $\mathbb{Q}$.

Is this a valid example, to show the incompleteness of the rational numbers?

Mauro ALLEGRANZA
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some_user
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    Yes. Sometimes the example chosen when that statement is made is $\sum_{n=0}^\infty\frac1{n!}$, because the irrationality is easy to prove, but any rational sequence converging to an irrational number is a valid example. – Jean-Claude Arbaut Nov 28 '22 at 08:46
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    @Jean-ClaudeArbaut Even more classic is the sequence (not series) $$\frac{\left\lfloor 10^n\cdot\sqrt2\right\rfloor}{10^n}$$where the irrationality proof is even easier (or at least more universally known and celebrated). – Arthur Nov 28 '22 at 08:55
  • Yes, it is. As others have mentioned though, the digital expansion of any irrational number is too. (Arthur's is example is simply taking the digital expansion of $\sqrt 2$ and looking at it term by term. I can see how a novice student might think that that is cheating or begging the question but it really isn't. The subtle thing is, it is because the rationals are incomplete but dense, that we know we can write any real as a decimal.) – fleablood Nov 28 '22 at 09:16
  • You don't even need to know the irrationality of a named number to do this. Just use any decimal expansion that is not eventually periodic, such as $0.101001000100001 \ldots$ (continue increasing by one the number of zeros between consecutive $1$'s). No such number is rational because the decimal expansion of any quotient of nonzero integers is eventually periodic, a result that is straightforward from usual grade school long division and the pigeonhole principle. – Dave L. Renfro Nov 28 '22 at 09:24

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