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cherrilyn

  • 5 years ago

PROOF! of the sum a sub n........

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  1. cherrilyn
    • 5 years ago
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    Prove that if \[\sum_{}^{}a _{n} \] converges absolutely then \[\sum_{}^{} a _{n}^{2} \] also converges. Then show by giving a counterexample that \[\sum_{}^{} a _{n}^{2} \] need not converge if \[\sum_{}^{}a _{n} \] is only conditionally convergent.

  2. watchmath
    • 5 years ago
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    How rigorous do you want the proof to be? Is it using epsilon-delta type of proof?

  3. cherrilyn
    • 5 years ago
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    no..we just learned about harmonic, p-series, alternating harmonic...absolute and conditional convergence....... so I don't think that rigorous

  4. watchmath
    • 5 years ago
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    Well, we can do the following: Since the first series is convergent, then there is an \M\) such that \(0\leq |a_n|<1\) for all \(n\geq M\) (think that if there are infinitely many terms that greater than 1, then the series diverges). But then \(a_n^2<a_n\) for all \(n\geq M\) Hence by the comparison test the second series is convergent as well.

  5. watchmath
    • 5 years ago
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    Example: \[\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}\] is is conditionally convergent but \[\sum_{n=1}^\infty \frac{1}{n}\]is divergent.

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