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May I know the answer for this with respect to

  1. Jun 26, 2011 #1
    May I know the answer for this with respect to t:

    [tex]\sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan(\frac{2t}{4n+1}))[/tex]

    Thanks in advance.
     
    Last edited: Jun 26, 2011
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  3. Jun 26, 2011 #2

    hunt_mat

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    Re: Summation((t/2n)-arctan(2t/(4n+1))

    Can I just expand the series as:
    [tex]
    \frac{t}{2}\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\tan^{-1}\frac{2t}{4n+1}
    [/tex]
    and the first term is infinite unless t=0.
     
  4. Jun 26, 2011 #3

    micromass

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    Re: Summation((t/2n)-arctan(2t/(4n+1))

    No, that won't work. A difference of two divergent series may be divergent. A simple example is

    [tex]\sum_{n=1}^{+\infty}{\frac{1}{n}-\frac{1}{n}}[/tex]

    I can show that the series of the OP converges using the integral test. But it seems that the OP wants the exact value of the series. I have no clue how to calculate that...
     
  5. Jun 26, 2011 #4
    Re: Summation((t/2n)-arctan(2t/(4n+1))

    I have an idea, but I don't know where to go with it yet; try finding an expression whose derivative or integral is t/2n - arctan(2t/(4n+1)) and make that a function of t, and try to solve with power series manipulations.
     
  6. Jun 26, 2011 #5

    disregardthat

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    Re: Summation((t/2n)-arctan(2t/(4n+1))

    There is no such thing as a difference between two divergent series. (sum 1/n) - (sum 1/n) and sum (1/n -1/n) are two very different things (the first one does not make sense at all).

    The equivalence (sum a_n) + (sum b_n) = sum (a_n+b_n) is true (and makes sense) if and only if the sums exists (that is, converges).
     
  7. Jun 28, 2011 #6
    Re: Summation((t/2n)-arctan(2t/(4n+1))

    Using a calculator, I got infinity
     
  8. Jun 28, 2011 #7
    Re: Summation((t/2n)-arctan(2t/(4n+1))

    Writing t/(2n)-arctan(2*t/(4n+1)) as a series in t and then summing each coefficient as a series in n (in Maple) gives
    (2-3/2*ln(2)-1/4*Pi)*t-1/48*Psi(2,5/4)*t^3+1/3840*Psi(4,5/4)*t^5-1/645120*Psi(6,5/4)*t^7+1/185794560*Psi(8,5/4)*t^9+O(t^11)
    Psi(n,x) is the nth derivative of Psi(x), so it seems possible that the series may be expressible in terms of Psi(x). At least, this method might give you a reasonable approximation to the original series. I haven't thought carefully about the convergence issues, but I believe the original series does converge.
     
  9. Aug 8, 2011 #8
    Re: Summation((t/2n)-arctan(2t/(4n+1))

    t is a constant.

    Anyway, I realised it should converge. It is related to the Riemann-Siegel Theta function.

    [tex]\theta (t) = - \frac{t\, \gamma + t\;\mbox{ln}\: \pi}{2}-\arctan 2t + \sum_{n=1}^{\infty}(\frac{t}{2n}-\arctan\frac{2t}{4n+1})[/tex]

    and this theta function does converge, at least when the Z-function is concerned.
     
  10. Aug 8, 2011 #9
    Re: Summation((t/2n)-arctan(2t/(4n+1))

    Put

    [tex]\arctan\frac{2t}{4n+1} = \int_0^t \frac{2(4n+1)}{(4n+1)^2+4x^2}dx[/tex]

    and interchanging the sum and integral and using Mathematica gives

    [tex]\int_0^t \sum_{n=1}^{\infty} \frac{1}{2n} -\frac{2(4n+1)}{(4n+1)^2+4x^2} dx
    = \frac{\gamma t}{2} + \frac{1}{2i}(\log\Gamma(5/4+it/2)-\log\Gamma(5/4-it/2))[/tex]

    It should be possible to put the solution in real form, for example with the right integral representation of log(Gamma).
     
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