Convergence and Cauchy Criterion

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SUMMARY

The sequence (xn) defined by the condition |xn + 1 - xn| < 1/n² is proven to be convergent using the Cauchy criterion. The proof involves showing that for any given ε > 0, there exists an integer N such that for all m > n > N, the inequality |xn - xm| < ε holds. The calculations demonstrate that the differences between terms in the sequence can be bounded effectively, confirming convergence. However, clarity in defining ε as a positive real number and ensuring the proof structure is logically sound is essential for completeness.

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Homework Statement


Suppose the sequence (xn) satisfies |xn + 1 - xn| < 1/n2, prove that (xn) is convergent.

Homework Equations


|xn - xm| < ɛ

The Attempt at a Solution


If m > n, then
|xn - xm|
< |xn - xn + 1| + |xn + 1 - xn + 2| + ... + |xm - 1 - xm|
< 1/n2 + 1/(n+1)2 + ... + 1/(m - 1)2
< [1/(n - 1) - 1/n] + [1/n - 1/(n + 1)] + ... + [1/(m - 2) - 1/(m - 1)] = 1/(n - 1) - 1/(m - 1)
< 1/(n - 1)

Let ɛ be given. Choose m > n > N := [1/ɛ] + 1 such that |xn - xm| < ɛ for all m > n > N.

Is there any problem with my proof?
 
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The calculations look good, but the statements surrounding it do not. In particular:
1. It doesn't make sense to say that you choose m such that some statement is true for all m.
2. You didn't say that ##\varepsilon## is a positive real number, so the reader can wonder how you intend to make ##|x_n-x_m|<\varepsilon## when ##\varepsilon=-1##.
3. The calculation didn't actually conclude that ##|x_n-x_m|<\varepsilon##.
Edit: 4. If the [x] notation means what I think it does, your definition of N doesn't work.
 
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