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Sum or upped bound of geometrico-harmonic series

  1. Jun 20, 2010 #1
    Hi,
    I need help to determine the upper bound of this infinite series.
    [tex]\sum_{k=p+1}^{\infty} \frac{1}{k} a^k \ \ \ \ ; a \leq 1[/tex]

    The paper I am reading reports the upper bound to be,
    [tex]\sum_{k=p+1}^{\infty} \frac{1}{k} a^k \leq \frac{1}{p+1}\sum_{k=p+1}^{\infty} a^k = \frac{1}{p+1} \cdot \frac{a^{p+1}}{1-a}[/tex]

    I totally cannot understand how could he factor 1/k coefficient out of the summation and replace it with 1/(p+1). Please point me in the right direction.
     
    Last edited: Jun 20, 2010
  2. jcsd
  3. Jun 20, 2010 #2

    CRGreathouse

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    k is at least p+1, so 1/k is at most 1/(p+1).
     
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