Results 1 to 7 of 7

Math Help - [SOLVED] Partial fractions for series?

  1. #1
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1

    [SOLVED] Partial fractions for series?

    Yep, it's me again....

    \sum^{\infty}_{n=2} \frac{2}{n^2 - 1}

    Should I use partial fractions on this one, or is there a better way to find the sum?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Yep. It's a telescoping series.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mollymcf2009 View Post
    Yep, it's me again....

    \sum^{\infty}_{n=2} \frac{2}{n^2 - 1}

    Should I use partial fractions on this one, or is there a better way to find the sum?
    to find the sum, partial fractions is a good way to go (think "telescoping series"). or use algebra:

    \frac 2{(n + 1)(n - 1)} = \frac {n + 1 - n + 1}{(n + 1)(n - 1)} = \frac 1{n - 1} - \frac 1{n + 1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    to find the sum, partial fractions is a good way to go (think "telescoping series"). or use algebra:

    \frac 2{(n + 1)(n - 1)} = \frac {n + 1 - n + 1}{(n + 1)(n - 1)} = \frac 1{n - 1} - \frac 1{n + 1}

    Ok, so when I work this out, I am left with:

    \lim_{n \rightarrow \infty} 1 - \frac{1}{n+1}  = 1

    This does not seem right to me.

    How high do i need to evaluate this in terms of n? I mean in terms of cancelling out terms?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by mollymcf2009 View Post
    Ok, so when I work this out, I am left with:

    \lim_{n \rightarrow \infty} 1 - \frac{1}{n+1}  = 1

    This does not seem right to me.

    How high do i need to evaluate this in terms of n? I mean in terms of cancelling out terms?
    nope, that's not right. you are missing some guys. for instance, 1/2 doesn't cancel. check again


    you need to evaluate the first n terms

    try the following just to see a pattern:

    1st term + 2nd term + 3rd term + 4th term + 5th term + ...... + (n - 3)th term + (n - 2)th term + (n - 1)th term + nth term

    where you see the dots, you skip those, just jump from the 5th term to the (n - 3)th term. our goal is to see the pattern of what cancels and what stays. if you are keen on picking up the pattern early, you might not have to test that many
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member mollymcf2009's Avatar
    Joined
    Jan 2009
    From
    Charleston, SC
    Posts
    490
    Awards
    1
    Quote Originally Posted by Jhevon View Post
    nope, that's not right. you are missing some guys. for instance, 1/2 doesn't cancel. check again


    you need to evaluate the first n terms

    Ok, got it! 3/2 Thanks so much for all of your help y'all! I REALLY appreciate it!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    \frac{1}{n-1}-\frac{1}{n+1}=\left( \frac{1}{n-1}-\frac{1}{n} \right)+\left( \frac{1}{n}-\frac{1}{n+1} \right), and those are telescoping series.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sum of a series using partial fractions.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 6th 2010, 03:44 AM
  2. Replies: 0
    Last Post: April 28th 2010, 09:53 AM
  3. [SOLVED] help partial fractions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 2nd 2009, 10:09 PM
  4. Taylor Series partial fractions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 8th 2008, 04:41 PM
  5. [SOLVED] Partial Fractions #1
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 6th 2008, 12:19 PM

Search Tags


/mathhelpforum @mathhelpforum