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by RiderOfGiraffes
5610 days ago
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As a simple counter-example to your claim, the sequence: x, x^x, x^x^x, x^x^x^x, ...
when x = sqrt(2) is strictly increasing and bounded above, and therefore converges. It's not hard to show that it's bounded above by 2, because x^x^2 > x^x^x, and x^x^2 = x^2 = 2. Repeat for any length sequence of exponentiation. |
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