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we've learned about the Bolzano-Weierstrass theorem that states that if a sequence is bounded, then it has a subsequence that converges to a finite limit. Let's define $a_n$ as the digits of $\pi$, i.e. $a_1$ = 3, $a_2$ = 1, $a_3$ = 4, and so on infinitely. Certainly this sequence is bounded by 10 and 0, but I can't think of any subsequence that will converge to anything. Can you help solve my confusion?

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    $\begingroup$ Just because you can't explicitly give an example does not mean one doesn't exist. $\endgroup$ – David Mitra Dec 13 '14 at 13:31
  • $\begingroup$ But, see this. $\endgroup$ – David Mitra Dec 13 '14 at 13:33
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Sure, take the subsequence of every occurrence of $1$. If there aren't infinitely many $1$s, then $2$s; if not $2$s then $3$s, etc. As $\pi$ is irrational, at least one non-zero digit in the decimal expansion is repeated infinitely many times.

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    $\begingroup$ This works with every number, it is not required that the number is irrational. The digits can be 0... so some digit will repeat infinitely many times. $\endgroup$ – Emanuele Paolini Dec 13 '14 at 13:41
  • $\begingroup$ Maybe he was thinking of two digits (possibly including 0) occurring infinitely often; then irrationality comes into play. $\endgroup$ – Vandermonde Dec 13 '14 at 23:04
  • $\begingroup$ It's a matter of convention: I usually think of a decimal expansion as finite "if it's finite", e.g., 1.2, not 1.200000.... $\endgroup$ – Simon S Dec 14 '14 at 3:34

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