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Math Help - Need Help with Series!

  1. #1
    Newbie
    Joined
    Mar 2009
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    Need Help with Series!

    hey guys, really need help with calculus... its my last class i need to pass in order to gaduate and get
    the dammed bachelors degree. any who...

    the problem is that i dont understand anything i started doing the homework i need to finish today .. and got stuck on the first exercise here it is..

    series of (3/n^2 - 16) n = 0 to infinity
    <br /> <br />
\sum^{\infty}_{n=0} ({\frac{{\frac{3}{(n+1)^2 -16}}}{{\frac{3}{n^2 - 16}}}})<br />

    ok i know this is a Series P and P = 2 therefore is greater than 1 making this series converge

    thing is i need to apply the ratio formula.. wich is :

    limit(3/(((n+1)^2-16)*(3/(n^2-16))), n = infinity)

    maple says the limit is 1, but when i do it manually i get -(12/15)

    what am i doing wrong??


    here are my steps

    3
    ---------
    (n+1)^2 -16
    -----------
    3
    ---------
    n^2 - 16

    ok so ...the 3's cancel each other out and you get

    lim (n+1)^2 -16
    ----------
    n^2 - 16

    now you make a substitution, n = 1

    lim (1+1) ^2 - 16
    --------------
    1^2 - 16

    that turns into

    lim 4 - 16
    ------
    1 - 16

    that equals

    - 12/15

    now... -12/15 is not 1....


    what am i doing wrong?? thx in advance!!
    Last edited by vipersnake; March 24th 2009 at 06:38 AM. Reason: added the graphical equation, any wasy way to make this formulas? ;)
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  2. #2
    Super Member Showcase_22's Avatar
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    I turned it into a_n=\frac{n^2-16}{(n+1)^2-16}

    From here it's apparent that a_n \rightarrow 1 (and therefore the series does not converge since a_n \rightarrow 0 for that to happen) but you need the ratio test.

    Hence:

    |\frac{a_{n+1}}{a_n}|=\frac{((n+1)^2-16)((n+1)^2-16)}{((n+2)^2-16)(n^2-16)} \rightarrow 1 (you can expand all this out if you want, it isn't apparent it converges to 1 in this form).

    I don't know why you substituted 1 into the equation.

    Is this what you needed?
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  3. #3
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    Showcase_22, thank you so much for your fast reply, basically i need to know whether it converges or not using the ratio, direct comparison and the integral test.

    and i needed to know how the limit of
    <br />
a_n=\frac{n^2-16}{(n+1)^2-16}<br />

    got to 1, since in class the profeesor told us that if the lim was 1 it was inconclusive but if the limit as less than 1 it converged and if it was greater than 1 then it diverged.
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  4. #4
    Super Member Showcase_22's Avatar
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    With the ratio test, it's |\frac{a_{n+1}}{a_n}| \rightarrow L as n \rightarrow \infty not as n \rightarrow 1.

    Since the ratio test tends to 1, we can't use it here (as you rightly said).

    The series does not converge. A necessary condition for it to converge is that a_n \rightarrow 0 but it doesn't. Isn't this enough?

    If not, try and make \sigma a_n larger than something that diverges.
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