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Sum of sequence

  1. Jul 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Let {a_n} be the sequence defined by
    [tex]a_{n} = \frac{n^{2}+8n+10}{n+9}[/tex]

    Find the value of [tex]\sum_{n=1}^{30}\ a_n[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]\sum_{n=1}^{30}\ a_n[/tex]

    = [tex]\sum_{n=1}^{30}\ n -1+\frac{19}{n+9}[/tex]

    = [tex]\sum_{n=1}^{30}\ n - [/tex] [tex]\sum_{n=1}^{30}\ 1 +[/tex][tex]\sum_{n=1}^{30}\frac{19}{n+9}[/tex]

    = [tex]1/2 (30)(31) - 30 + [/tex] [tex]\sum_{n=1}^{30}\frac{19}{n+9}[/tex]

    I'm stuck here...

    thx
     
    Last edited: Jul 25, 2009
  2. jcsd
  3. Jul 25, 2009 #2

    tiny-tim

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    Hi songoku! :smile:

    (you mean (n + 9) :wink:)

    Hint: can you do [tex]\sum_{n=1}^{39}\frac{1}{n}[/tex] ? :smile:
     
  4. Jul 25, 2009 #3
    Hi tiny-tim :)

    Oh yes i mean n+9 :wink:

    Sorry i don't know how to do that....

    more hint?

    thx
     
  5. Jul 25, 2009 #4

    tiny-tim

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    Hi songoku! :wink:

    ok … new hint … can you sum [tex]\sum_{n=1}^{39}x^n[/tex] and then integrate it? :smile:
     
  6. Jul 25, 2009 #5
    Hi tiny-tim :tongue2:

    Are you trying to say that :

    [tex]\sum_{n=1}^{30}\frac{19}{n+9} = \int_{0}^{30}\frac{19}{n+9} dn ?[/tex]

    I did try your last hint and got 1/40 x^40 + 1/39 x^39 + ... + 1/2x^2 after the integration...
     
  7. Jul 25, 2009 #6

    tiny-tim

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    Hi songoku! :biggrin:

    i] nooo …

    ii] try putting x = 1 :wink:
     
  8. Jul 25, 2009 #7
    Hi tiny-tim !:rofl:

    I laughed a lot when i read this. I'm pretty sure that you won't state that [tex]\sum_{n=1}^{30}\frac{19}{n+9} = \int_{0}^{30}\frac{19}{n+9} dn [/tex]

    but i just gave it a shot and posted it, lol

    1/2 + 1/3 + ... + 1/40 ???

    i'm still trying to catch the hint :biggrin:
     
    Last edited: Jul 25, 2009
  9. Jul 25, 2009 #8

    tiny-tim

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    Hi songoku! :wink:

    Sorry … I've just realised my method doesn't work for reciprocals :redface:

    (I was thinking that the sum is (1 - xn)/(1 - x), which I could then integrate, and put x = 1, but there isn't an easy way to integrate it :rolleyes:)

    In fact, I don't think there is any "short-cut" solution for this …

    I think you have to use the "psi function" (see http://en.wikipedia.org/wiki/Digamma_function#Recurrence_formula)
     
  10. Jul 25, 2009 #9
    Hi tiny-tim ! :biggrin:

    OMG psi function....

    I think maybe i'll do it manually...

    thx a lot tiny-tim ^^
     
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