Results 1 to 5 of 5

Math Help - Telescoping Series

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    46

    Telescoping Series

    a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}

    It is known that limit of the terms is 0 and that the series converges to \frac{-1}{6}.

    I have a problem that asks for c_n where a_n = c_n - c_{n+1}.

    Maybe it is simple, but I'm having trouble envisioning how to find c_n. If anyone has any tips or hints, I'd greatly appreciate it.

    Thanks for your consideration,

    -Austin Martin
    Follow Math Help Forum on Facebook and Google+

  2. #2
    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 auslmar View Post
    a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}

    It is known that limit of the terms is 0 and that the series converges to \frac{-1}{6}.

    I have a problem that asks for c_n where a_n = c_n - c_{n+1}.

    Maybe it is simple, but I'm having trouble envisioning how to find c_n. If anyone has any tips or hints, I'd greatly appreciate it.

    Thanks for your consideration,

    -Austin Martin
    do you know how to find partial fractions?

    \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)} = - \frac 12 \sum^{\infty}_{n=1} {\color{red} \frac{1}{(n+2)(n+3)}}

    can you find the partial fractions for what is in red? do that, then write out the first few terms, then the last few terms, as many as you need to see a pattern of what would cancel out. then take the limit as n \to \infty of what's left
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2008
    Posts
    46
    Quote Originally Posted by Jhevon View Post
    do you know how to find partial fractions?

    \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)} = - \frac 12 \sum^{\infty}_{n=1} {\color{red} \frac{1}{(n+2)(n+3)}}

    can you find the partial fractions for what is in red? do that, then write out the first few terms, then the last few terms, as many as you need to see a pattern of what would cancel out. then take the limit as n \to \infty of what's left
    Oh yes, I've already found the answer to that. I'm just having trouble identifying " c_n."
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by auslmar View Post
    a_n = \sum^{\infty}_{n=1}\frac{-1}{2(n+2)(n+3)}

    It is known that limit of the terms is 0 and that the series converges to \frac{-1}{6}.

    I have a problem that asks for c_n where a_n = c_n - c_{n+1}.

    Maybe it is simple, but I'm having trouble envisioning how to find c_n. If anyone has any tips or hints, I'd greatly appreciate it.

    Thanks for your consideration,

    -Austin Martin
    \frac{1}{(n+2)(n+3)}=\frac{1}{n+2}-\frac{1}{n+3}

    \therefore\quad{c_n=\frac{1}{n+2}}

    I assume that this is what you are asking
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Just for kicks, do you want to see the telescope?. Perhaps you already know it, but here is how they look at it and get -1/6 in the event you don't.

    Using \frac{1}{2(n+3)}-\frac{1}{2(n+2)}

    \left(\frac{1}{8}-\frac{1}{6}\right)+\left(\frac{1}{10}-\frac{1}{8}\right)+\left(\frac{1}{12}-\frac{1}{10}\right)+\left(\frac{1}{14}-\frac{1}{12}\right)+.............

    As you can see, all the terms cancel one another out except the -1/6.

    See?.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: April 13th 2011, 12:39 PM
  2. Telescoping series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 21st 2010, 04:18 AM
  3. Telescoping Series
    Posted in the Calculus Forum
    Replies: 9
    Last Post: April 15th 2010, 09:58 AM
  4. Telescoping Series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 30th 2009, 10:28 AM
  5. Telescoping series
    Posted in the Calculus Forum
    Replies: 7
    Last Post: March 25th 2009, 01:20 AM

Search Tags


/mathhelpforum @mathhelpforum