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So I'm typing on a phone or else I'd use math Jax, but basically take an infinite sequence that is made of the values of the square root function as each successive term. This clearly does not converge to a real number. Yet it is a Cauchy Sequence of real numbers. How can this be? Bare in mind I didn't reach very long for duplicates, and I just learned of Cauchy sequences literally today in class.

Christheyankee
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  • "Infinite sequence that is made of the values of the square root function..." Elaborate on this. So do you mean something like $\sqrt{n}$, or $\sqrt{1 + {1\over n}}$? – Decaf-Math Nov 20 '18 at 21:28
  • I'm not sure I understood the definition of the sequence. You are taking square root of which numbers? Anyway, if you got that a Cauchy sequence of real numbers does not converge to a real number then clearly you did something wrong. – Mark Nov 20 '18 at 21:29
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    If you mean the sequence obtained by iterating square root on some iniitial value: $x, \sqrt{x}, \sqrt{\sqrt{x}}, \ldots$, then the sequence does converge for non-negative real $x$. (And if asked to prove that this sequence is Cauchy, I'd do it by proving that it converges $\ddot{\smile}$.) – Rob Arthan Nov 20 '18 at 21:31
  • @Christheyankee The sequence is Not cauchy, the difference between two CONSECUTIVE terms goes to 0, but not the difference between 2 arbitrary terms both after a fixed N – Anshuman Agrawal Mar 18 '23 at 09:23

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