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Math Help - Limits of series

  1. #1
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    Limits of series

    having trouble finding limits of series,

    I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

    SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

    im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

    Any guidance here would be greatly appreciated,

    Cheers.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by monster View Post
    having trouble finding limits of series,

    I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

    SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

    im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

    Any guidance here would be greatly appreciated,

    Cheers.
    Hint: Almost all of these kind of sums are telescoping. Use partial fractions and see what pops out.
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  3. #3
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    Quote Originally Posted by monster View Post
    having trouble finding limits of series,

    I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

    SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

    im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

    Any guidance here would be greatly appreciated,

    Cheers.
    Try using partial fractions...

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

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


    Therefore A(4n + 1) + B(4n - 3) = 1

    4An + A + 4Bn - 3B = 1

    (4A + 4B)n + A - 3B = 0n + 1.


    Therefore 4A + 4B = 0 and A - 3B = 1.

    From the first equation we can see A = -B.

    Substituting into the second equation

    -B - 3B = 1

    -4B = 1

    B = -\frac{1}{4}.

    So A = \frac{1}{4}.



    So we can write

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


    Therefore

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

     = \frac{1}{4}\left[\sum_{n = 1}^{\infty}\left(\frac{1}{4n - 3}\right) - \sum_{n = 1}^{\infty}\left(\frac{1}{4n + 1}\right) \right]

     = \frac{1}{4}\left[\left(1 + \frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \frac{1}{17} + \dots\right) - \left(\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \frac{1}{17} + \dots\right)\right]

     = \frac{1}{4}\left(1\right)

     = \frac{1}{4}.
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