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Math Help - Dirichlet Series convergence

  1. #1
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    Dirichlet Series convergence

    Hey guys, while studying dirichlet series I came along a theorem stating that every D.S. has a convergence abscissa (possibly +Infinity). Now there was a corollary stating that any Ordinary D.S. \sum_{n=1}^{\infty} a_n/n^s has a absolute convergence abscissa, which doesn't differ by more than 1 from the normal abscissa, but there is no proof. Can anyone prove this?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by EinStone View Post
    Hey guys, while studying dirichlet series I came along a theorem stating that every D.S. has a convergence abscissa (possibly +Infinity). Now there was a corollary stating that any Ordinary D.S. \sum_{n=1}^{\infty} a_n/n^s has a absolute convergence abscissa, which doesn't differ by more than 1 from the normal abscissa, but there is no proof. Can anyone prove this?
    For those who aren't familiar with Dirichlet series but might be able to help, could you fill in some of the terminology.
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    MHF Contributor chisigma's Avatar
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    According to...

    Dirichlet series - Wikipedia, the free encyclopedia

    ... a D.S. of the form \varphi (s)= \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} converges in the following cases...

    a) if a_{n} is a bounded sequence of complex numbers for Re(s)>1 ...

    b) if a_{n} = \mathcal{O} (n^{k}) for Re(s)> k+1...

    c) if \sum_{i=n}^{n+k} a_{i} is bounded forall n and k\ge 0 for Re(s)>0...

    Kind regards

    \chi \sigma
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    A Dirichlet series is just a series \sum_{n=1}^{\infty} \dfrac{a_n}{n^s}, where a_n is any sequence of complex numbers, and s\in \mathbb{C}. We consider the convergence of the series depending on s.

    I need help with the proof of a corollary, stating that if \sum_{n=1}^{\infty} \dfrac{a_n}{n^{s_0}} converges for some s_0 \in \mathbb{C}, then it converges absolutely for any s_1 \in \mathbb{C} with Re(s_1) > Re(s_0)+1
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    Quote Originally Posted by EinStone View Post
    A Dirichlet series is just a series \sum_{n=1}^{\infty} \dfrac{a_n}{n^s}, where a_n is any sequence of complex numbers, and s\in \mathbb{C}. We consider the convergence of the series depending on s.

    I need help with the proof of a corollary, stating that if \sum_{n=1}^{\infty} \dfrac{a_n}{n^{s_0}} converges for some s_0 \in \mathbb{C}, then it converges absolutely for any s_1 \in \mathbb{C} with Re(s_1) > Re(s_0)+1
    It looks simple: since the series for s_0 converges, the general term \frac{a_n}{n^{s_0}} tends to 0.

    However, \left|\frac{a_n}{n^s}\right|=\left|\frac{a_n}{n^{s  _0}}\right|\frac{1}{|n^{s-s_0}|} and |n^z|=|e^{z\log n}|=e^{{\rm Re}( z)\log n}=n^{{\rm Re} (z)}. You have every element you need to conclude!
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    I see n^{{\rm Re} (z)} converges to 0, for n -> Infinity, but the other term is still in absolute value, so how do you show convergence now, with Cauchy criterion?
    Last edited by EinStone; February 19th 2010 at 10:29 AM.
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    Quote Originally Posted by EinStone View Post
    I see n^{{\rm Re} (z)} converges to 0, for n -> Infinity, but the other term is still in absolute value, so how do you show convergence now, with Cauchy criterion?
    Because of my first point, there is C>0 such that we have \left|\frac{a_n}{n^{s_0}}\right|\leq C for all n\geq 1 (sequence converges to zero hence, in particular, is bounded). Then \left|\frac{a_n}{n^s}\right|\leq \frac{C}{n^{{\rm Re}(s-s_0)}}, and we know well that \sum_n \frac{C}{n^{{\rm Re}(s-s_0)}} converges when {\rm Re}(s-s_0)>1 (Riemann series). This implies that \sum_n \left|\frac{a_n}{n^s}\right| converges, which by definition is the same as saying that \sum_n \frac{a_n}{n^s} converges absolutely.

    I notice that you first mentioned normal convergence, so here it is (almost the same proof): let s_1 be such that {\rm Re}(s_1)>{\rm Re}(s_0)+1. We have to prove that the series converges normally on the set \{z\in\mathbb{C}|{\rm Re}(z)>{\rm Re}(s_1)\}. This is just because, for every s is this set, we have \left|\frac{a_n}{n^s}\right|\leq \frac{C}{n^{{\rm Re}(s_1-s_0)}} and \sum_n \frac{C}{n^{{\rm Re}(s_1-s_0)}}<\infty.
    Last edited by Laurent; February 19th 2010 at 01:25 PM.
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  8. #8
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    Thats it, thanks!
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