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Math Help - Riemann Zeta function

  1. #1
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    Riemann Zeta function

    Hey, so following problem:

    I want to express (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}

    as a Dirichlet series, i.e. as a series of the form

    \sum_{n=1}^{\infty}\frac{a_n}{n^s} where a_n is any complex sequence.

    Who can help?
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  2. #2
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    Quote Originally Posted by EinStone View Post
    Hey, so following problem:

    I want to express (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}

    as a Dirichlet series, i.e. as a series of the form

    \sum_{n=1}^{\infty}\frac{a_n}{n^s} where a_n is any complex sequence.

    Who can help?

    I'm not sure I follow you, but what if you define a_n=1-2^{1-s}\,\,\forall n\in\mathbb{N} ? This certainly will do the trick...

    Now perhaps you're on to something else, since (1-2^{1-s})\zeta(s) is the first step to express the zeta function as an Euler product, so...well, so who knows what you're on to actually.

    Tonio
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  3. #3
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    Quote Originally Posted by EinStone View Post
    Hey, so following problem:

    I want to express (1-2^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s}

    as a Dirichlet series, i.e. as a series of the form

    \sum_{n=1}^{\infty}\frac{a_n}{n^s} where a_n is any complex sequence.

    Who can help?
    i have a good reason to believe that a_n=(-1)^{n+1} also works but i need some time to prove it. EDIT: it's actually very easy to prove!
    Last edited by NonCommAlg; February 23rd 2010 at 05:02 AM.
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  4. #4
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    Quote Originally Posted by tonio View Post
    I'm not sure I follow you, but what if you define a_n=1-2^{1-s}\,\,\forall n\in\mathbb{N} ? This certainly will do the trick...

    Now perhaps you're on to something else, since (1-2^{1-s})\zeta(s) is the first step to express the zeta function as an Euler product, so...well, so who knows what you're on to actually.

    Tonio
    I'm sorry I didnt clarify this, a_n should not depend on s or anything else, otherwise its trivial ^^.


    Quote Originally Posted by NonCommAlg View Post
    i have a good reason to believe that a_n=(-1)^{n+1} also works but i need some time to prove it. EDIT: it's actually very easy to prove!
    I also found that result online, but I don't see how to prove this , can you help?
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  5. #5
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    \sum_{n \geq 1} \frac{(-1)^{n+1}}{n^s}=\sum_{n \geq 1} \frac{1}{(2n-1)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}=\sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}

    =\sum_{n \geq 1} \frac{1}{n^s} -2\sum_{n \geq 1} \frac{1}{(2n)^s}= \sum_{n \geq 1} \frac{1}{n^s} -2^{1-s} \sum_{n \geq 1} \frac{1}{n^s}=(1-2^{1-s}) \sum_{n \geq 1} \frac{1}{n^s}.
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    \sum_{n \geq 1} \frac{1}{(2n-1)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}=\sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} - \sum_{n \geq 1} \frac{1}{(2n)^s}

    I understand everything except the quoted one above, so why is

    \sum_{n \geq 1} \frac{1}{(2n-1)^s} = \sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} ?
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  7. #7
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    Quote Originally Posted by EinStone View Post
    I understand everything except the quoted one above, so why is

    \sum_{n \geq 1} \frac{1}{(2n-1)^s} = \sum_{n \geq 1} \frac{1}{n^s} - \sum_{n \geq 1} \frac{1}{(2n)^s} ?


    \sum_{n \geq 1} \frac{1}{(2n-1)^s}=\frac{1}{1^s}+\frac{1}{3^s}+\frac{1}{5^s}+\l  dots=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+\ld  ots -\left(\frac{1}{2^s}+\frac{1}{4^s}+\ldots\right)

    Tonio
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  8. #8
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    Ah too easy xD.

    So I have another really similar problem. Namely expressing
    (1-3^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s} as a Dirichlet series.

    I tried again to split up the n on three series for the modulo 3 residues, but it somehow didnt work. Do you have an idea?


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  9. #9
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    Quote Originally Posted by EinStone View Post
    Ah too easy xD.

    So I have another really similar problem. Namely expressing
    (1-3^{1-s})*\sum_{n=1}^{\infty}\frac{1}{n^s} as a Dirichlet series.

    I tried again to split up the n on three series for the modulo 3 residues, but it somehow didnt work. Do you have an idea?

    a_n=-2 \cos \left(\frac{2 \pi n}{3} \right) works.

    in general, let d \in \mathbb{N} and define f : \mathbb{N} \longrightarrow \{0,1\} by f(n)=\begin{cases}1 & \text{if} \ d \mid n \\ 0 & \text{otherwise} \end{cases}. let a_n=1-df(n). then it's easy to prove that \sum_{n \geq 1} \frac{a_n}{n^s}=(1-d^{1-s})\sum_{n \geq 1} \frac{1}{n^s}.
    Last edited by NonCommAlg; February 24th 2010 at 10:36 AM.
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