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Series summation

  1. Sep 30, 2006 #1

    I'm working on a little puzzle and part of it requires summing the infinite series 1/(k^1.5) which clearly converges, but I've never been very good at actually finding what a series converges to. Could you give me a good swift kick in the head. Just a hint will do.

  2. jcsd
  3. Sep 30, 2006 #2
    does it converge to something like 2.61216453017413....? thats what i got by adding the first 90 million terms.
  4. Sep 30, 2006 #3
    Yeah I know, but how to show to what it converges exactly?
  5. Sep 30, 2006 #4
    Here's something I've been trying. bear with me



    is the laplace transform of

    \sqrt{t} \frac{2}{\sqrt{\pi}}

    So we can rewrite our series as

    \sum_{k=1}^{\infty} \frac{1}{k^{1.5}} = \sum_{k=1}^{\infty} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sqrt{t} e^{-kt} dt

    The integral is pretty nice being positive and decreasing so there's probably a nice theorem out there saying we can swap sum and integral, which gives us

    \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \sqrt{t} \sum_{k=1}^{\infty} e^{-kt} dt

    The sum is a geometric one, or pretty close, and we get the following

    \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{\sqrt{t}e^{-t}}{1-e^{-t}} dt

    Now I'm a big baby and can't figure out how to integrate this. I shove it in mathematica (and replace infinity with a big number) and I get the same (approximately) as I do when evaluate the truncated series. Which is about 2.61238. Now I'm sure there's a nice way to evaluate this integral and get an exact answer. Any takers?
    Last edited: Sep 30, 2006
  6. Sep 30, 2006 #5


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    Well, you have the first two decimal places correct. It's 2.6123753486854883433485675679240716305708006524....

    It's just the zeta function, which is easy to calculate.
  7. Sep 30, 2006 #6
    Well I looked into the link and I can see lots of information on the zeta function evaluated at integer values but nothing about fractional powers like 3/2. Could you outline how to get an exact expression for something like this?


  8. Sep 30, 2006 #7
    it converges to [itex]\zeta (\frac{3}{2})[/itex]. What makes you think there's a nicer way to write it down? :smile:
  9. Sep 30, 2006 #8


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    There are several mentions of fractional values, just none for 3/2. I doubt there's a closed form expression; finding closed-form expressions for zeta values other than even integers is hard in general.
  10. Sep 30, 2006 #9
    I only thought it converged to a nicer expression because its part of a puzzle I was working on. You are given a gift that is unusually wrapped. It is shaped like a stack of cubes, the first one is 1 foot hight the second is [itex]
    The third is
    and so on. So the height is clearly infinite. The problem also asks about the surface area, which is infinites and the volume which gives us the series I started with, which is finite. Given that it was a puzzle, I sort of supposed that it would come out nicer. Thanks for all the input.
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