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Sum of a series

  1. May 9, 2012 #1
    Find the sum of the series:

    Summation (n=1 to infinity) 2^(1/n)

    I already know that this is convergent. I don't see this as a geometric series so that means I have to use partial sums. And I know that the sum is equal to the limit of the sequence of partial sums. I just don't know how to don't out, there's not good enough examples in the book that go along with this
  2. jcsd
  3. May 9, 2012 #2
  4. May 9, 2012 #3


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    It is divergent as the term sequence converges to 1, and not 0.
  5. May 9, 2012 #4
    Oh yea I lied it is divergent. But I don't know why it is divergent either haha. How does it converge to 1? The terms are 2 + sqrt(2) + 2^(1/3) + 2^(1/4)+...
  6. May 9, 2012 #5
    Just a guess, here. I'm still trying to learn these type of series.

    Since the term at n = infinity converges to 1, the sum would continue for ever, 1 being added at every successive ∞ + 1, ∞ +2, so on. Therefore you would never get a exact limiting sum, which makes it divergent.

    I believe this is correct, though it would be great if someone confirmed it.
  7. May 9, 2012 #6
    Definitely makes sense, I didn't think of it that way!
  8. May 9, 2012 #7

    Ray Vickson

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    The series is divergent, and you do not need powerful software to tell you that. The nth term a_n= 2^(1/n) does not approach 0 as n approaches infinity.

  9. May 9, 2012 #8
    So when I look for convergence or divergence of a series, if it doesn't approach 0, it's automatically divergent? And then all other numbers are convergent? What are like the 'rules' to it
  10. May 9, 2012 #9

    Ray Vickson

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    If the nth term does not --> 0 the series is divergent; if the nth term does --> 0 the series may converge or it may diverge: additional work is needed to tell which occurs.

  11. May 10, 2012 #10
    what kind of additional work? also, this is the first section on series, meaning we havent been taught any of the convergence tests...
  12. May 11, 2012 #11


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    Using the tests that you'll be soon seeing.
  13. May 11, 2012 #12


    Staff: Mentor

    What Ray is talking about here is something you'll see very soon - the n-th term test for divergence. It is pretty important, but deceptively simple, saying that if the limit of the n-th term is not 0, then the series it's part of diverges.

    It doesn't say anything at all about series whose n-th term has a limit of zero.
  14. May 11, 2012 #13
    i feel lost with the book, it doesn't show how to determine if a series (without it being geometric) converges or diverges but then for every question they ask if the series converges or diverges. its confusing
  15. May 11, 2012 #14


    Staff: Mentor

    I think you are missing something that is in your book. If they expect you to determine convergence/divergence, they must have given you some tools to work with. Take a closer look at any examples and theorems they present in that section.
  16. May 11, 2012 #15
    The Ratio Test and the Integral Test are always nice.
  17. May 11, 2012 #16
    we didnt learn the ratio test or integral test yet. but from this post and another post of mine, i think i may have it
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