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Series converging to zero?

  1. Dec 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi all, I've been banging my head at this one for a while so any help is appreciated!

    Consider the series,
    [tex]\frac{1}{n}\sum_{k = 1}^{n} \frac{1}{k^{\delta}}[/tex]
    where [tex]\delta \in (0,1)[/tex]. My teacher tells me that this series converges to zero but I am unable to prove that it is so.


    2. Relevant equations



    3. The attempt at a solution
    Indeed, I can easily show that this series converges (i.e. to something, but not necessarily to zero) by doing the following. Since for [tex]k = 1, 2, \ldots, n[/tex], we have that [tex]k \leq n[/tex], then it folloWs that [tex]k^{\delta} \cdot k = k^{1 + \delta} \leq k^{\delta}n[/tex], and this implies we have [tex]\frac{1}{k^{1 + \delta}} \geq \frac{1}{k^{\delta}n}[/tex]. And since [tex]1 + \delta > 1[/tex], then it implies [tex]\sum \frac{1}{k^{1 + \delta}}[/tex] is a convergent p-series and hence by the Comparision Test, it follows that [tex]\lim_{n \to \infty} \sum_{k = 1}^{n} \frac{1}{n} \frac{1}{k^{\delta}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k = 1}^{n} \frac{1}{k^{\delta}} < \infty[/tex].

    --- but then again, what I need is to show that the series not only converges, but converges to zero! (note that if we had [tex]1/n^2[/tex], instead of just [tex]1/n[/tex], then this problem immediately follows from what I'd shown above. Thanks!
     
  2. jcsd
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