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Series summation

  1. Sep 30, 2006 #1
    hello,

    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.

    Thanks,
     
  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

    Now

    [itex]
    \frac{1}{(k^{1.5})}
    [/itex]

    is the laplace transform of

    [itex]
    \sqrt{t} \frac{2}{\sqrt{\pi}}
    [/itex]

    So we can rewrite our series as

    [itex]
    \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
    [/itex]

    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

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

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

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

    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

    CRGreathouse

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    Science Advisor
    Homework Helper

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

    It's just the zeta function, which is easy to calculate.
    http://mathworld.wolfram.com/RiemannZetaFunction.html
     
  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?

    Thanks

    Kevin
     
  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

    CRGreathouse

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    Science Advisor
    Homework Helper

    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]
    \frac{1}{\sqrt{2}}
    [/itex]
    The third is
    [itex]
    \frac{1}{\sqrt{3}}
    [/itex]
    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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