Results 1 to 4 of 4

Thread: partial sum question

  1. September 9th 2010, 04:20 AM #1
    newcastle
    newcastle is offline
    Newbie
    Joined
    Sep 2010
    Posts
    1

    partial sum question

    hey guys i'm having a little trouble working out a question. the question is to find a formula for the nth partial sum of the series sum of 1/((4n+1)(4n-3)) between n=1 and infinity. any help would be great thanks
    Follow Math Help Forum on Facebook and Google+
  2. September 9th 2010, 04:22 AM #2
    yeKciM
    yeKciM is offline
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    454
    Quote Originally Posted by newcastle View Post
    hey guys i'm having a little trouble working out a question. the question is to find a formula for the nth partial sum of the series sum of 1/((4n+1)(4n-3)) between n=1 and infinity. any help would be great thanks
    first use partial fractions to write that sum than write few of them (depending how many you need to realize which is formula for nth partial sum)

    show what have you done



    \displaystyle s = \sum _{n=1} ^{N} \frac {N}{4N+1}
    Follow Math Help Forum on Facebook and Google+
  3. September 9th 2010, 06:45 AM #3
    chisigma
    chisigma is offline
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Is...

    \displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (\frac{1}{4n-3} - \frac{1}{4n+1}) (1)

    ... so that...

    \displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{4}\ (1 - \frac{1}{4N+1}) (2)

    ... and finally...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4} (3)

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 9th 2010 at 08:14 AM. Reason: 7 instead 4... error signalled by yeKciM... sorry...
    Follow Math Help Forum on Facebook and Google+
  4. September 9th 2010, 06:52 AM #4
    yeKciM
    yeKciM is offline
    Senior Member yeKciM's Avatar
    Joined
    Jul 2010
    Posts
    454
    Quote Originally Posted by chisigma View Post
    Is...

    \displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (\frac{1}{4n-3} - \frac{1}{4n+1}) (1)

    ... so that...

    \displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{7}\ (1 - \frac{1}{4N+1}) (2)

    ... and finally...

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7} (3)

    Kind regards

    \chi \sigma


    probably just typo it's

    \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4} (3)





    Edit:


    \displaystyle \frac {1}{(4n+1)(4n-3)} = \frac {A}{4n+1} + \frac {B}{4n-3}

    4A+4B =0 \Rightarrow A=-B

    \displaystyle -3A+B=1 \Rightarrow -4A=1 \Rightarrow A=-\frac {1}{4} \Rightarrow B = \frac {1}{4}

    ...... ........
    Follow Math Help Forum on Facebook and Google+
« radioactive substance problem | Factoring an expression that results from using the "product rule" »

Similar Math Help Forum Discussions

  1. ono-one function question and partial fractions question
    Posted in the Algebra Forum
    Replies: 12
    Last Post: October 2nd 2011, 06:07 AM
  2. Partial Fractions Question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 4th 2010, 04:49 PM
  3. Partial differentiation question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 25th 2009, 02:23 PM
  4. A partial fraction question..
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 22nd 2008, 09:27 AM
  5. partial fraction question
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: November 23rd 2006, 06:10 AM

Search Tags


/mathhelpforum @mathhelpforum