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Math Help - Sum of two series equals zero

  1. #1
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    Sum of two series equals zero

    I've been messing around with the plotter from the site:
    Dirichlet Series Plotter

    And it seems like any imaginary part put into the equation is a reflection about the x-axis of the negative imaginary part. I need someone to cofirm the equation in the attachment because I don't think this is adding up correctly.
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  2. #2
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    Quote Originally Posted by rman144 View Post
    I've been messing around with the plotter from the site:
    Dirichlet Series Plotter

    And it seems like any imaginary part put into the equation is a reflection about the x-axis of the negative imaginary part. I need someone to cofirm the equation in the attachment because I don't think this is adding up correctly.
    Assuming b \in \mathbb{R}:

    Well the imaginary part of:

     \sum_{n=1}^{\infty}\frac{1}{n^{1/2-ib}} +\sum_{n=1}^{\infty}\frac{1}{n^{1/2+ib}}

    if it exists, is zero, since the partial sums of the first series are equal to the conjugates of the partial sums of the second. However since you cannot rearrange conditionally convergent series what I say below about the real part also applies here.

    The real part is:

    2 \sum_{n=1}^{\infty}\frac{\cos(b\ln(n))}{n^{1/2}}

    If this is zero (or even converges, since it does not satisfy the conditions of the alternating serier test) I don't know.

    (The integral does not converge, but that also means nothing as the conditions for the integral test for convergence are not satisfied by this series)

    If I were a betting person I think I would put my money on non-convergence.

    (Note b=0 is divergent)

    It might be interesting to consider other definitions of convergence for this problem. (for some reason I cannot find any references on line for these at present),
    but I can givew an example:

    For the dubiously convergent series \sum_{n=1}^{\infty} a_n , define the "sum" to be:

    S^*=\lim_{\alpha \to 0} \sum_{n=1}^{\infty} a_n e^{-\alpha n}

    RonL
    Last edited by CaptainBlack; July 3rd 2008 at 12:22 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    It might be interesting to consider other definitions of convergence for this problem. (for some reason I cannot find any references on line for these at present),
    but I can givew an example:

    For the dubiously convergent series \sum_{n=1}^{\infty} a_n , define the "sum" to be:

    S^*=\lim_{\alpha \to 0} \sum_{n=1}^{\infty} a_n e^{-\alpha n}

    RonL
    Further online research after a bit of lateral thinking about search strings give this reference.

    The above example appears to be equivalent to Abel summation

    RonL
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  4. #4
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    I know it is unusual to answer my own question, but the resulting series diverges. Test a few values on the site:

    On-Line Calculator

    And you will see that the values progressively grow. I don't have a rigid proof, but the numbers would indicate a divergent series.
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  5. #5
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    Quote Originally Posted by rman144 View Post
    I know it is unusual to answer my own question, but the resulting series diverges. Test a few values on the site:

    On-Line Calculator

    And you will see that the values progressively grow. I don't have a rigid proof, but the numbers would indicate a divergent series.
    Well I can't get that calculator to sum series properly for me, but the sum:

    <br />
S_N=\sum_{n=1}^{N}\frac{\cos(\ln(n))}{n^{1/2}}<br />

    does not grow without bound as N \to \infty it ossilates with increasing amplitude (at least for the first 10,000,000 terms).

    RonL
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  6. #6
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    I think you may have made a typo there. The equation you just posted is missing an imaginary component and would thus be zeta(1/2). My equation was zeta(1/2+bi). You are correct however in saying the equation oscillates. However, although it oscillates, it diverges to greater value with each oscillation (on average).
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