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Convergence of Series

  1. Nov 12, 2005 #1
    Our instructor assigned a problem from Rudin's Principles of Mathematical Analysis; a problem which I have been unable to solve after giving it good thought.

    The statement is:
    "Prove that the convergence of SUM[an] implies the convergence of SUM[sqrt(an)/n], if an >= 0."

    The instructor did give us a hint: "Review the ideas of chapter one," from which I gleaned the archimedean property or supremums might be important. Anyway, if anyone is familiar with how this can be proved, I would appreciate a nudge in the right direction. Thanks.
     
  2. jcsd
  3. Nov 12, 2005 #2
    In case anyone else read this, I have found the solution on the internet. It was embarassingly simple. By the AM-GM inequality:

    [tex]a_n + \frac{1}{n^2} \ge 2\frac{\sqrt{a_n}}{n}.[/tex]​

    The left hand series converges, so by direct comparison, the right hand series also converges.
     
    Last edited: Nov 12, 2005
  4. Nov 12, 2005 #3

    quasar987

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    Cool, I didn't know of this inequality. What does AM-GM stands for?
     
  5. Nov 13, 2005 #4

    benorin

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    Harmonic - Geometric - Arithmetic Mean Inequality

    AM-GM = Arithmetic Mean - Geometric Mean

    In full, the Harmonic - Geometric - Arithmetic Mean inequality is (for sequences)

    [tex]\frac{n}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdot\cdot\cdot +\frac{1}{a_{n}}}\leq \left( a_{1}a_{2}\cdot\cdot\cdot a_{n}\right) ^{\frac{1}{n}}\leq \frac{a_{1}+a_{2}\cdot\cdot\cdot + a_{n}}{n}[/tex]

    where equality holds iff the [itex]a_{i}[/itex]'s are all equal, and it is understood that [itex]a_{i}\geq 0,\forall i[/itex].
     
  6. Nov 13, 2005 #5

    benorin

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    Harmonic - Geometric - Arithmetic Mean inequality is (for functions)

    Suppose that the real-valued function [tex]f(x)[/tex] is defined, properly integrable, and strictly positive on the interval [tex] \left[ x_{1}, x_{2}\right] [/tex].

    Then the Harmonic - Geometric - Arithmetic Mean inequality is (for functions)

    [tex] \frac{x_{1} - x_{2}}{\int_{x_{1}}^{x_{2}} \frac{dx}{f(x)}} \leq \exp\left( \frac{1}{x_{1} - x_{2}}\int_{x_{1}}^{x_{2}} \log f(x) dx\right) \leq \frac{1}{x_{1} - x_{2}} \int_{x_{1}}^{x_{2}} f(x)dx[/tex]
     
    Last edited: Nov 13, 2005
  7. Nov 14, 2005 #6

    quasar987

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    Great, I'm noting all of this down.
     
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