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Proving a sequence is a Cauchy Sequence

  1. Apr 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove [itex]\sum\frac{(-1)^k}{k^2}[/itex] is a Cauchy sequence.


    2. Relevant equations
    Definition of Cauchy sequence: |a[itex]_{n}[/itex] - a[itex]_{m}[/itex]|<ε for all n,m>=N, n>m

    3. The attempt at a solution

    I thought if I could prove that the above summation was less than the summation of 1/k^2, the problem would be easy. But either that isn't true, or I don't know how to prove it.

    So I started some stuff with the triangle inequality. Here's what I have in full so far:

    Let ε>0 be given. Choose a positive integer N such that 1/N<ε. Then, if m,n>=N, and n>m, we have |a_n - a_m| <= |[itex]\sum(-1)^n/n^2[/itex]| + |-[itex]\sum(-1)^m/m^2[/itex].

    But I don't really know where to go from here. I could say that |a_m| = |-a_m|, but I don't think that gets me anywhere. Any hints to get me going again would be greatly appreciated.
     
  2. jcsd
  3. Apr 19, 2012 #2
    It should be intuitively clear that [itex]|a+b| \leq |a| + |b|[/itex]. From this follows that [itex]|\sum f(x)| \leq \sum |f(x)|[/itex], thus [itex]\sum \frac{(-1)^k}{k^2} \leq \sum \frac{|(-1)^k|}{k^2} = \sum \frac{1}{k^2}[/itex].
     
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