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Show the series (1/2)^sqrt(n) converges

  1. Apr 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Show [itex]\displaystyle\sum\limits_{n=0}^\infty (\frac{1}{2})^{\sqrt{n}}[/itex] converges.

    3. The attempt at a solution
    I've pretty much tried all convergence tests and failed miserably. Both the ratio and the root test are inconclusive, as the limits are 1. I can't really use the integral test, and I don't know what else to compare this to to show that the series indeed converges. I know it does, but I just can't show it.

    One idea I had is to rewrite the sum as [itex]\displaystyle\sum\limits_{n=0}^\infty (\frac{1}{2})^{\sqrt{n}} = \displaystyle\sum\limits_{n=0}^\infty (\frac{1}{2^{\frac{1}{\sqrt{n}}}})^{n}[/itex], where the thing in brackets is less than 1 for all n. Knowing the geometric series converges for r < 1, this would imply this series also converges... Except that I don't think reasoning is any good, since for any such r < 1, the thing in brackets will eventually get bigger than r. This is all I've got now, though.

    Ugh, any help here would be greatly appreciated, as always.
     
  2. jcsd
  3. Apr 9, 2012 #2

    SammyS

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    Try the integral test, or the comparison test.
     
  4. Apr 9, 2012 #3
    Hmm, how would I use the integral test? Integrate (1/2)√x, and show that the sum from 2 to n is smaller than the integral from 1 to n?

    As for the comparison test, like I mentioned, I did try it, I just couldn't find anything to compare the function to.
     
  5. Apr 9, 2012 #4
    If I am not mistaken, it is a theorem that if the integral method works, then the sum converges in all cases, so you wouldn't need to prove the sum is smaller than the sum.
     
  6. Apr 9, 2012 #5

    SammyS

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    A possible comparison ... It's a little complicated, but here's
    the idea. Look at the terms in which n is a perfect square.

    (Write (1/2)√n as 2-√n to simplify things.)

    For n = 1,2,3: 2-√1 ≥ 2-√n . There are 3 of these.

    For n = 4, … 8: 2-√4 ≥ 2-√n . There are 5 of these.

    For n = 9, … 15: 2-√9 ≥ 2-√n . There are 7 of these.

    For n = 16, … 24: 2-√16 ≥ 2-√n . There are 9 of these.

    See the pattern?
     
  7. Apr 9, 2012 #6

    Dick

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    (1/2)^sqrt(x) is a positive decreasing function so you can use the integral test by just showing the integral from some point to infinity is finite, like SammyS also suggested. You can integrate it. It's just an integration by parts after you substitute x=u^2.
     
    Last edited: Apr 9, 2012
  8. Apr 9, 2012 #7
    Heh, yeah, in the time in-between your posts, this approach is actually the one I had taken, but it's good to see a confirmation that it is the right one. Basically, if I didn't mess it up, it should hold that [itex]\displaystyle\sum\limits_{n=0}^\infty (\frac{1}{2})^{\sqrt{n}} \leq \displaystyle\sum\limits_{n=0}^\infty (\frac{1}{2})^{n} + \displaystyle\sum\limits_{n=0}^\infty (2n+1)(\frac{1}{2})^{n}[/itex], right?

    Ah alright, when first thinking about the integral test at all, I think the particular integral just seemed too daunting too approach, but I see that perhaps next time I should just start plugging and chugging. I do like the comparison approach better, though :)

    Good to know, although I probably couldn't use that directly, seeing as how we haven't mentioned it in class.
     
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