- #1
tylerc1991
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Homework Statement
Show that the sequence of real numbers defined by [itex]x_{n + 1} = x_n + \frac{1}{x_n^2}, \, x_1 = 1[/itex] is not a Cauchy sequence.
Homework Equations
A sequence [itex]\{ p_n \}[/itex] is Cauchy if and only if, for all [itex]\varepsilon > 0[/itex], there exists an [itex]N > 0[/itex] such that [itex]d(p_n, p_m) < \varepsilon[/itex] for all [itex]m, n > N[/itex].
The Attempt at a Solution
We can assume that [itex]d[/itex] is the usual metric on [itex]\mathbb{R}[/itex]. I don't even see where to begin. I see that the sequence is monotonically increasing, so that
[itex]1 = \frac{1}{x_1} > \frac{1}{x_2} > \frac{1}{x_3} > \dotsb.[/itex]
So
[itex]1 = \frac{1}{x_1^2} > \frac{1}{x_2^2} > \frac{1}{x_3^2} > \dotsb.[/itex]
To me it looks like the sequence is in fact Cauchy. Please help!