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I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."

Wolfram has this simple continued fraction for $\zeta(5):$ $[1;27,12,1,1,15,...]$. But we don't know if $\zeta(5)$ is rational or not, so I understood something wrong.

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    $\begingroup$ I think WA is just giving the first few terms of the simple continued fraction. $\endgroup$ – saulspatz Jan 6 at 4:16
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The problem is we don't know what the "$\ldots$" is. If it terminates after finitely many terms, $\zeta(5)$ is rational. If it doesn't, $\zeta(5)$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.

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