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One quick question

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the series n= 1 to infinity of ( 1/n ^5 ) correct to three decimal places .
    2. Relevant equations

    We have to find Sn by sub. n number of terms


    3. The attempt at a solution

    But , we are not given n, so how to find the sum in such a case, is there a way to find n ?

    ( I mean by n the number of Sn terms )

    any help would be nice
     
  2. jcsd
  3. Apr 4, 2009 #2
    Any help will be great ..
     
  4. Apr 4, 2009 #3

    Pengwuino

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    You are given n, n=1 to infinity. The summation is defined as the sum of in this case, an infinite number of terms of [tex]\frac{1}{{n^5 }}[/tex]. If the problem stated n = 1 to 3, for example, it would simply be [tex]\frac{1}{{1^5 }} + \frac{1}{{2^5 }} + \frac{1}{{3^5 }}[/tex]
     
  5. Apr 4, 2009 #4
    Do you mean that I should make an integral n=1 to infinity ?

    even without telling that n=1 ?
     
  6. Apr 4, 2009 #5
    Ya, I tried that and I got .25
    and what I did is that I itegrated the given function and then evaluated the integtal
    with replacing the infinity with t and this what I got..

    So " it seems that we are not using the Sn , partial sums at all" Right ?
     
  7. Apr 4, 2009 #6

    Pengwuino

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    No, n is an index, not a variable in the usual sense. n counts from 1, 2, 3, 4, to infinity in integer steps. It sounds like an integral but it's completely different.
     
  8. Apr 4, 2009 #7
    Yes, sure I understand all what you said,
    But still did not answer my question,

    Is the way I did correct ,, integration without any Sn ??
     
  9. Apr 4, 2009 #8

    Pengwuino

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    There is no integration involved, you're simply adding [tex] \frac{1}{{1^5 }} + \frac{1}{{2^5 }} + \frac{1}{{3^5 }}[/tex] all the way to n = infinity. Now the question wants you to do this until the first 3 digits no longer change when you continue to add.
     
  10. Apr 4, 2009 #9
    I was thinking that the anwer should be rounded to 3 decimal places
     
  11. Apr 4, 2009 #10
    Because it saya " correct to three decimal places " ??

    What do you think ??
     
  12. Apr 4, 2009 #11

    Pengwuino

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    If you start calculating each term, you'll quickly realize that the first 3 decimal places no longer change in the sum after a certain n. That means you've found the sum accurate to 0.001.
     
  13. Apr 4, 2009 #12
    Uha, but this is not always
    It happenes to be correct in this case

    Right ?
     
  14. Apr 4, 2009 #13

    Pengwuino

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    As long as the series isn't divergent, each additional sum will add smaller and smaller amounts to the sum. However, I don't believe that the digits are guaranteed to stay the same as your summation approaches infinity.
     
  15. Apr 4, 2009 #14
    Ok, and when it says accurate to three decimal places that means

    0.000x

    where x is any number, I mean to say that three decimal places means three zeros
     
  16. Apr 4, 2009 #15

    Dick

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    Why are you messing around with this? Why don't you use the remainder estimate Tom Mattson suggested in your other thread? Rn is less than or equal to an integral.
     
  17. Apr 4, 2009 #16
    Uha, am ok with that, but I was wondering about the meaning of correct to three decimal places mean ?
     
  18. Apr 4, 2009 #17

    Dick

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    I think they probably just mean make sure that Rn<0.001. Making sure that really means you have 3 correct decimal places could be a little tricky if you have, let's say Sn=0.1234 and Rn<0.001. That really doesn't tell you exactly how to round off. But I don't think they mean to be tricky.
     
  19. Apr 4, 2009 #18
    Still, I don't feel that my question is anwered..

    What does "THREE DECIMAL PLACES MEAN"

    how many zeros after the point ??
     
  20. Apr 4, 2009 #19
    This is exactly what I mean ....
    I wish I could an "exact" answer .. before my exam tommoro
     
  21. Apr 4, 2009 #20

    Dick

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    Three places after the decimal point.
     
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