Results 1 to 5 of 5

Math Help - Finding bounds for a series

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    102

    Finding bounds for a series

    Use the integral test to find upper and lower bounds on the limit of the series:

     \sum_{j=1}^{\infty} je^{-j}

    So do I first turn this series into an integral? So this would then be...

     \int_{1}^{\infty} je^{-j}

    Is this right so far? If so, I'm stuck right here.
    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 larson View Post
    Use the integral test to find upper and lower bounds on the limit of the series:

     \sum_{j=1}^{\infty} je^{-j}

    So do I first turn this series into an integral? So this would then be...

     \int_{1}^{\infty} je^{-j}

    Is this right so far? If so, I'm stuck right here.
    integral test is way too much trouble. but since they asked for it...

    yes, that is ok. do integration by parts on this guy. let u = j and dv = e^{-j}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Jhevon View Post
    integral test is way too much trouble. but since they asked for it...

    yes, that is ok. do integration by parts on this guy. let u = j and dv = e^{-j}
    So...  u = j ... \frac {du}{dx} = 1 ... du = dx

    Then...  dv = e^{-j}dx ... \frac {dv}{dx} = e^{-j} ... v = \frac {e^{-j}}{-1}

    Now it gets messy...

    \int{u}dv=uv-\int{v\cdot{du}}

     j\frac {e^{-j}}{-1} - \int{\frac {e^{-j}}{-1}dx}

    Is this right as of now?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    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 larson View Post
    So...  u = j ... \frac {du}{dx} = 1 ... du = dx

    Then...  dv = e^{-j}dx ... \frac {dv}{dx} = e^{-j} ... v = \frac {e^{-j}}{-1}

    Now it gets messy...

    \int{u}dv=uv-\int{v\cdot{du}}

     j\frac {e^{-j}}{-1} - \int{\frac {e^{-j}}{-1}dx}

    Is this right as of now?
    let's not make our notation make the problem more complicated than it already is. \frac {e^{-j}}{-1} = - e^{-j}

    so your integral is: -je^{-j} + \int e^{-j}~dj

    and you can take it from there. start plugging in the limits. of course, we have to use limits for the infinite integration limit
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2008
    Posts
    102
    Quote Originally Posted by Jhevon View Post
    let's not make our notation make the problem more complicated than it already is. \frac {e^{-j}}{-1} = - e^{-j}

    so your integral is: -je^{-j} + \int e^{-j}~dj

    and you can take it from there. start plugging in the limits. of course, we have to use limits for the infinite integration limit
    So this then equals  -je^{-j} + \frac{e^{-j}}{-1} which of course means...

     -je^{-j} - e^{-j} so now just plug in the limits, aka 1 and \infty

    And after you do that I get... 2e^{-1} which is the lower bound, then to find the upper one wouldn't it just be... 3e^{-1}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding Upper and Lower Bounds
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 28th 2010, 05:59 PM
  2. Finding integral upper and lower bounds
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 15th 2010, 02:13 AM
  3. Bounds of the Harmonic Series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 15th 2009, 12:00 PM
  4. Finding the bounds for double integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2008, 06:49 PM
  5. Help finding bounds
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 24th 2008, 01:22 PM

Search Tags


/mathhelpforum @mathhelpforum