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Sum of Series

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate the sum of the following:

    [itex]\sum\limits^\inf_{n=1} \frac{6}{n(n+1)}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    Well... The denominator is going to get infinitely large as n approaches infinity, so would the value of the sum not converge to zero? The homework program said this answer was wrong.
     
  2. jcsd
  3. Nov 15, 2011 #2

    LCKurtz

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    The first term is 3 and they are all positive, so how could the sum be zero?

    Hint: Use partial fractions
    [tex]\frac 6 {n(n+1)}= \frac A n + \frac B {n+1}[/tex]
    and write out the first n terms.
     
  4. Nov 15, 2011 #3

    Ray Vickson

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    So, by your reasoning the sum S = 1/2 + 1/4 + 1/8 + 1/16 + ... = Ʃ 1/2n is zero, because the terms are going to zero. Does that look right to you?

    RGV
     
  5. Nov 15, 2011 #4
    Alright, I've separated the equation into the following:

    [itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

    And solved for A, B, getting A = 6, B = 0.

    So, I'm left with:

    [itex]\sum\limits^\infty_{n=1} \frac{6}{n}[/itex]

    Now, I'm lost as to what to do. I don't know how to solve for this sum...

    Any hints?
    Thank you so much PhysicsForums!
     
  6. Nov 15, 2011 #5

    Dick

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    Your partial fraction calculation is obviously wrong. 6/n isn't equal to 6/(n*(n+1)). Try and notice things like that! Try it again. Show your work if you can't get it right.
     
  7. Nov 15, 2011 #6
    Alright, well this is what I've done:

    [itex]\frac{6}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}[/itex]

    [itex]\frac{6}{n(n+1)} = \frac{A(n+1) + Bn}{n(n+1)}[/itex]

    [itex]6 = A(n+1) + Bn[/itex]

    [itex]6 = An + A + Bn[/itex]

    Split into two equations,

    [itex]1: 6 = A + B[/itex]
    [itex]2: 6 = A[/itex]

    Then,

    [itex]2->1: 6 = 6 + B[/itex]
    [itex]B = 0[/itex]

    I can see I'm doing something wrong, because [itex]\frac{6}{n}[/itex] does not equal [itex]\frac{6}{n(n+1)}[/itex].... What am I doing wrong?
     
  8. Nov 15, 2011 #7

    LCKurtz

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    Before you split it into two equations write the 3rd equation above like this:

    0n + 6 = (A+B)n + A

    So what should you get for A+B by equating coefficients?
     
  9. Nov 15, 2011 #8

    Dick

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    Your first equation should be 0=A+B not 6=A+B. The n's need to cancel if you are always going to get 6.
     
  10. Nov 15, 2011 #9
    Ah, perfect. I love you guys, you're so helpful. I guess I need to be a little more critical of my own work though, before I give up on what I've done completely.

    Now though, knowing A & B (I guess I was solving for them wrong), I am still confused as to how to get the 'nth term' of a series.

    So I found that A=6, B=-6. Then, I'm looking for

    [itex]\sum\limits^\infty_i=0 \frac{6}{n} - \frac{6}{n+1}[/itex]

    [itex] = \sum\limits^\infty_i \frac{6}{n} - \sum\limits^\infty_i \frac{6}{n+1}[/itex]

    I don't know how to get the formula for the nth term of these sums, but I do know how to solve once I get that formula. Can someone give me a push?

    Don't worry, I'm not just waiting for help from you guys.. I'm searching for tutorials on the Internet as well.. I missed my class on series / sums and don't know what I'm doing ... Thank you so much..
     
  11. Nov 15, 2011 #10

    Ray Vickson

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    You should have [itex] \sum_{n=1}^\infty [ \frac{6}{n} - \frac{6}{n+1} ], [/itex] not what you wrote; go back and look carefully at what you wrote. Now just write down the sum for n going from 1 to N for a few small values of N, to see what is happening.

    RGV
     
  12. Nov 15, 2011 #11

    LCKurtz

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    Leave them grouped together and write out the first several terms and see if you notice anything.
     
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