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Sum to infinity

  1. May 17, 2017 #1
    1. The problem statement, all variables and given/known data
    evaluate ## \sum _{n=1}^{\infty }4^{\frac{1}{n}}-4^{\frac{1}{n+2}}## .
    fc981b864d95a636c4f08b9deb209cd6.png
    https://holland.pk/uptow/i4/fc981b864d95a636c4f08b9deb209cd6.png

    2. Relevant equations
    telescoping series: sum = infinite lim (a1-a(n+1))
    S=a/(1-r)

    3. The attempt at a solution
    as the latter function is of "n-2" instead of "n-1", it isn't telescoping series, right?
    however, I don't know how to prove it is divergent instead of convergent in this question.
    thank you very much!
     
  2. jcsd
  3. May 17, 2017 #2

    Ray Vickson

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    Why do you think it is divergent? Are you sure that Wolfram Alpha has not gotten it wrong?

    Try a few finite sums first, such as ##\sum_{n=1}^4 \cdots## or ##\sum_{n=1}^8 \cdots##. This will give you some valuable insights.
     
  4. May 17, 2017 #3
    Indeed, it is one of the past exam paper question which we are not allowed to use calculator.
    To calculate by hand would be far too time consuming in exam.
    Other than try out the values, is there any other ways to calculate it?
    Thanks.
     
  5. May 17, 2017 #4

    DrClaude

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    Strangely, I don't get the same answer as you do with exactly the same input on WolframAlpha.

    In any case, you should use parentheses, as in your case WolframAlpha considered only the first term as part of the sum.
     
  6. May 17, 2017 #5
    Thanks for telling me that. The new answer I got from WolframAlpha is 3.98619, about 4.
    How can I get it by hand?
    Thanks
     
  7. May 17, 2017 #6

    Ray Vickson

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    You can do what I suggested in post #2. I never suggested doing calculations; I suggested writing down the sums, and all that is perfectly feasible in an exam without use of a calculator.

    Anyway, the Wolfram Alpha result is close, but not exact.
     
  8. May 17, 2017 #7
    But I need to have steps in long question... What I mean is an accurate answer in each steps...
    Calculating by trials isn't kind of valid steps, isn't it?
     
  9. May 17, 2017 #8

    Ray Vickson

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    Instead of arguing about it, why don't you just try what I suggested? I did not make the suggestion lightly, just to waste your time; I had a very good reason.
     
  10. May 17, 2017 #9
  11. May 17, 2017 #10

    WWGD

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    Like Ray Vickson suggested, rewrite the sum as a "sum of sums" and change the indices so that there is some cancellation.
     
  12. May 17, 2017 #11

    Ray Vickson

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    No, neither is correct.

    Have you done what I suggested in post #2?
     
  13. May 17, 2017 #12
    Yes, I did that.

    I found, ##\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + 4^{1/3} + ... ## And ##\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} = 4^{1/3} + 4^{1/4} + ... ##

    So the red part in ##\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... }## is ##\displaystyle \sum^\infty_{n = 1} 4^{1/(n+2)} ##.

    ##\displaystyle \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \color{red}{4^{1/3} + 4^{1/4} + ... } = \sum^\infty_{n = 1} 4^{1/n} = 4 + 4^{1/2} + \sum^\infty_{n = 1} 4^{1/(n+2)} \iff \\ \displaystyle \sum^\infty_{n = 1} 4^{1/n} - \sum^\infty_{n = 1} 4^{1/(n+2)} = 4 + 2 = 6 ##

    Why this wrong ?
     
  14. May 17, 2017 #13

    Mark44

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    Look at the partial sum, ##\sum_{n=1}^k \left( 4^{1/k} - 4^{\frac 1 {k+2}}\right)##. I find it helpful to write the positive terms in a column on the left and the terms being subtracted in a column on the right. There is a lot of telescoping going on, but possibly not as much as you think.

    Be sure to write the terms for k=1 through 6 or so, as well as the last 4 or 5 terms. Then take the limit as ##k \to \infty##. Remember that the sum of an infinite series is the limit of the sequence of partial sums as more and more terms are taken.
     
  15. May 17, 2017 #14

    Mark44

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    This is what I'm describing in the previous post.
    ##4^1~~~ -4^{1/3}##
    ##4^{1/2}~~~ -4^{1/4}##
    ##4^{1/3}~~~ -4^{1/5}##
    ##4^{1/4}~~~ -4^{1/6}##
    ##4^{1/5}~~~ -4^{1/7}##
    ##\vdots##
    Be sure to continue the pattern for the last three or four pairs of terms.
     
  16. May 17, 2017 #15
    ##\\ 4^{1/1} - 4^{1/3}\\ 4^{1/2} - 4^{1/4}\\ 4^{1/3} - 4^{1/5}\\~~~ \vdots \\ 4^{1/(n-2)} - 4^{1/(n)}\\ 4^{1/(n-1)} - 4^{1/(n+1)}\\ 4^{1/(n)} - 4^{1/(n+2)}\\ ##

    So sum to ##n## is ##4 + 4^{1/2} - 4^{1/(n+1)} - 4^{1/(n+2)}## Now I take limit n to infinity ??
     
  17. May 17, 2017 #16

    Mark44

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    Yes, and the limit as ##n \to \infty## is what?
     
  18. May 17, 2017 #17
    0 ?
     
  19. May 17, 2017 #18

    Mark44

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    No, but this is Yecko's problem. Don't do his work for him.
     
  20. May 17, 2017 #19
    Yes got it !!
     
  21. May 17, 2017 #20

    Ray Vickson

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    That is exactly what I suggested the OP do, but he refused to cooperate, so I gave up.
     
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