Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the value approximately of 1/1*2+1/2*3+1/3*4

  1. Jul 9, 2005 #1
    find the value approximately of
    1/1*2+1/2*3+1/3*4.........
    Is there a simple way to find the anwer to problems like this?????
     
  2. jcsd
  3. Jul 9, 2005 #2
    If i understand your question correctly, then
    1. do u know the sigma notation?
    2. can you write your series using the sigma notation?
    3. once you are done with step 2, can you do some simplification?
    4. do u know what is a telescoping series?

    -- AI
     
  4. Jul 9, 2005 #3
    Oh man, I thought it was 1/(1*2)+1/(2*3) ... and that was really hard. That's so much easier though.
     
  5. Jul 9, 2005 #4
    Hmmm... I guess it could be like that too. The OP will probably post if my interpretation was incorrect.
     
  6. Jul 9, 2005 #5

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    Erm, that is really easy to work out and if it isn't that what else could it be, after all:

    (1/1)*2 + (1/2)*3 + (1/3)*4 + ...

    Does clearly not converge.
     
  7. Jul 9, 2005 #6
    Doesn't converge? Would you explain why not? I see a convergence.
     
  8. Jul 9, 2005 #7

    LeonhardEuler

    User Avatar
    Gold Member

    The terms aren't even approaching zero.
     
  9. Jul 9, 2005 #8
    Right, but there is a cancellation.
     
  10. Jul 9, 2005 #9

    LeonhardEuler

    User Avatar
    Gold Member

    What are you talking about? Every term is greater than 1, so the sum is greater than the number of terms. Therefore as the number of terms approaches infinity, so does the sum.
     
  11. Jul 9, 2005 #10
    Woah, my bad guys. I totally looked at that the wrong way. Sorry about that. I'm tired.

    The series should be.... [tex]a_{n}=\frac{1}{n*(n+1)}[/tex]

    Correct?
     
    Last edited: Jul 9, 2005
  12. Jul 9, 2005 #11

    LeonhardEuler

    User Avatar
    Gold Member

    Yes, I think thats what it is. He said he needed an approximation, so do you think he should just integrate it?
     
  13. Jul 9, 2005 #12

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper
    2015 Award

    just adding up a few terms makes it look really easy. but maybe i am too optimistic with small evidence.
     
  14. Jul 9, 2005 #13

    Curious3141

    User Avatar
    Homework Helper

    What is the need for approximation ? It's easy to get an exact answer.

    1) Sigma notation
    2) Partial fraction decomposition, splitting up the sum into the difference of two sums
    3) Inspection and cancellation of a *lot* of terms (this is pertaining to Tenali's mention of telescoping series)
    4) The answer, the end.
     
    Last edited: Jul 9, 2005
  15. Jul 10, 2005 #14

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    If the sum is:

    [tex]\sum_{n=1}^{\infty} \frac{1}{n(n+1)}[/tex]

    Then the solution is really simple. But the poster hasn't posted what they know, so it's not like we can do it for them....
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find the value approximately of 1/1*2+1/2*3+1/3*4
  1. < or = to 1 (Replies: 5)

  2. -1 = 1 ? (Replies: 9)

  3. Find the mistake 1=-1 ? (Replies: 10)

  4. 1 = -1 ? (Replies: 7)

Loading...