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Evaulate the Sum

  1. Oct 24, 2012 #1

    Ʃ 1/(n(n+8))
    n=1

    So i used partial fractions and got (1/8)/(n) - (1/8)/(n+8)

    From there i pulled out the 1/8 so now my equation is



    (1/8) Ʃ (1/n)-(1/(n+8))
    n=1

    So from here do i just start doing like s1= (1/8)(1-1/9), s2=(1/8)(1/2-1/10)

    to find a value it converges to?
     
  2. jcsd
  3. Oct 24, 2012 #2

    Dick

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    The sum of 1/n=1+1/2+1/3+1/4+...
    The sum of 1/(n+8)=1/8+1/9+1/10+1/11+...
    If you think about the difference, a LOT of terms cancel.
     
  4. Oct 24, 2012 #3
    When i get to terms that cancel out, what do i do to solve thats where im confused. my teacher never explained exactly what to do
     
  5. Oct 24, 2012 #4

    Zondrina

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    Notice that the terms 1 + 1/2 + 1/3 + ... stick around while the others get canceled?
     
    Last edited: Oct 24, 2012
  6. Oct 24, 2012 #5
    Yes so i just sum all those multiply by (1/8) and thats it? Im just trying to make sense of my teachers notes.
     
  7. Oct 24, 2012 #6

    Dick

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    Yes, just sum the terms that don't cancel and divide by 8.
     
  8. Oct 25, 2012 #7

    Ray Vickson

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    If [tex] S(N) = \sum_{n=1}^N \left[\frac{1}{n} - \frac{1}{n+8} \right]
    = \sum_{n=1}^N \frac{1}{n} - \sum_{n=1}^N \frac{1}{n+8},[/tex] try writing out a few of the S(N) values for N larger than 8, say for N = 10, 11, 12, etc. That will give you a feel for what is happening, and then the final solution will be easier to see.

    RGV
     
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