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Finding the sum of the series

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the sum of the series:
    n=1∑(6)/((2n-1)(2n+1))
    2. Relevant equations



    3. The attempt at a solution

    S1=2
    S2=2+(6/15)
    S3=2+(6/15)+(6/35)

    This is the part where I get a little confused. It looks like the denominator is getting bigger... So does it approach infinity and diverge?
     
  2. jcsd
  3. Mar 11, 2014 #2
    Hi jdawg!

    I suggest decomposing into partial fractions.
     
  4. Mar 11, 2014 #3
    Ohh! Thanks! I'll give that a shot :)
     
  5. Mar 11, 2014 #4
    (6)/((2n-1)(2n-1))=(A/(2n-1))+(B/(2n+1)

    6=A(2n+1)+B(2n-1)

    6=A2n+A+B2n-B

    6=A-B A2n+B2n=0
    A=6+B

    (6+B)2n+B2n=0
    12n+B2n+B2n=0
    12n-2B2n

    I don't think I'm doing this right
     
  6. Mar 11, 2014 #5
    Correct so far.
    I can't understand this step. What have you done here?
     
  7. Mar 11, 2014 #6
    Oops! Sorry, ignore that step. Could you do this?:

    12n+B2n+B2n=0
    B(2n+2n)=-12n
    B(4n)=-12n
    B=-3

    A=6+B
    A=6-3
    A=3
     
  8. Mar 11, 2014 #7
    Looks right to me. :)

    Can you proceed with the problem?
     
  9. Mar 11, 2014 #8
    Awesome! Ok so:
    n=1 ∑((3)/(2n-1))-((3)/(2n+1))

    S1=2
    S2=2+(2/5)
    S3=2+(2/5)+(6/35)

    This is the part that really confuses me, how do you find the sum?

    Also, this might be a dumb question, but if you find what the series converges to is that number the sum, or is that just what the limit approaches? I don't have a very good grasp on series yet.
     
  10. Mar 11, 2014 #9
    Don't calculate the sums! Write the sums in the following way:
    $$S_1=\frac{3}{1}-\frac{3}{3}$$
    $$S_2=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}$$
    $$S_3=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}+\frac{3}{5}-\frac{3}{7}$$
    Do you see now? :)
     
  11. Mar 11, 2014 #10
    Ohhh! I see the pattern in the denominator now, it looks like its increasing by +2 each time?
    Is this the part where you do Sn?
    Sn=(3)/(n)-(3)/(n+2)
    I don't think that's quite right though...
     
  12. Mar 11, 2014 #11
    Not correct. If the individual sums doesn't help, you can write the given summation as follows:
    $$\sum_{n=1}^{\infty} \frac{3}{2n-1}-\frac{3}{2n+1}=3\left(\sum_{n=1}^{\infty} \frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
    $$=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7} \cdots \right)$$

    I hope this helps.
     
  13. Mar 11, 2014 #12
    Thanks so much, you were super helpful! :)
     
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