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Riemann Zeta approaches infinity as x approaches 1

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data


    We know [itex]\sum_{n=1}^{\infty}\frac{1}{n^x}[/itex] is uniformly convergent on the interval [itex]x\in(1,\infty)[/itex] and that its sum is called [itex]\zeta(x)[/itex]. Proof that [itex]\zeta(x) \rightarrow \infty[/itex] as [itex]x \rightarrow 1^+[/itex].


    2. Relevant equations

    We cannot find the formula that [itex]\zeta[/itex] is given by, but we know
    [itex]\sum_{n=1}^{\infty}\frac{1}{n^1}[/itex] diverges.

    3. The attempt at a solution
    I've already shown that the functional series is uniformly convergent on the above interval and that the derivative also converges uniformly to a limit function, thus [itex]\zeta(x)[/itex] is continuous and can be differentiated.

    The solution seem quite apparent: As [itex]x[/itex] approaches [itex]1[/itex], the series will approach the harmonic series, and thus it will diverge. Given any [itex]M>0[/itex], I can find a [itex]b>0[/itex] so that [itex]x \in (1,1+b) \Rightarrow \zeta(x) >M[/itex], as we know that we will approach the harmonic series.

    However, I'm not quite sure how I can go about formulating a formal proof. For functional series, I suppose I cannot just plug in the limit and say its a harmonic series (especially as it's not defined there).

    Under which conditions can I be completely sure that plugging in the relevant [itex]x[/itex] and then let [itex]n[/itex] approach infinity will be equivalent to letting n approach infinity and then let [itex]x[/itex] approach infinity? Or can I approach this in a different way?

    Any help will be greatly appreciated!
     
  2. jcsd
  3. May 17, 2012 #2
    Why not use the fact that
    $$
    \sum_{n=1}^{\infty}\frac{1}{n}
    $$
    is a harmonic series.

    Then
    $$
    1+\frac{k}{2}\leq\sum_{n=1}^{2^k}\frac{1}{n}
    $$
    as ##k\to\infty##
    $$
    \infty = 1+\frac{k}{2}\leq \sum_{n=1}^{2^k}\frac{1}{n}
    $$
     
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