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Triangle inequality with countably infinite terms

  1. Feb 12, 2009 #1
    In lecture in my real analysis course the other day we were proving that absolute convergence of a series implies convergence. Our professor started off by showing us the wrong way to prove it:

    [tex]\left| \sum_{k=1}^\infty a_k \right| \leq \sum_{k=1}^\infty \left| a_k \right| < \epsilon[/tex]

    Then he demonstrated the correct proof, by showing that the sequence of partial sums is Cauchy convergent and then using the triangle inequality.

    But this got me thinking: why is the first proof wrong? I definitely agree that the second proof is more solid, but if the triangle inequality is proved by induction, meaning it's true for all natural numbers, isn't that, well, infinite? I was wondering if someone could supply a counterargument or proof by contradiction illustrating why this conclusion is incorrect. Thanks as always.
     
  2. jcsd
  3. Feb 12, 2009 #2

    Office_Shredder

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    It's true for any natural number. Infinity isn't a natural number. That about sums it up.

    Obviously the conclusion that the triangle inequality holds for an infinite series holds, since you proved it did. But that doesn't mean the logic that you used to make that conclusion is necessarily correct.

    Here's an example that might elucidate the problem:

    Every sequence has an element of greatest magnitude. This is obviously true for finite sequences [tex]a_1, a_2,..., a_n[/tex] and is [tex]max_{i=1,...n}(| a_i |)[/tex]

    Hence, by your logic, if we have an infinite sequence [tex]a_1, a_2,... [/tex] then [tex]max_{i=1,2,...}( |a_i| )[/tex] exists and is in the sequence. Obvious counterexample: [tex]a_i = 1-\frac{1}{i}[/tex]
     
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