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Math Help - Converge of Diverge...why?

  1. #1
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    Converge of Diverge...why?

    Sum from 1 to infinity of:

    n/(1000n + 1) (does this converge or diverge...why?)


    My thought is take the lim from 1 to infinity of the above expression.

    That will end up 1/1000. So this converges...Right?
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  2. #2
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    Quote Originally Posted by jzellt View Post
    Sum from 1 to infinity of:

    n/(1000n + 1) (does this converge or diverge...why?)


    My thought is take the lim from 1 to infinity of the above expression.

    That will end up 1/1000. So this converges...Right?

    \sum_{n=1}^{\infty}\frac{n}{1000(n)+1}

    Note: The divergence test states if

    \lim_{n\to \infty} a_n \ne 0

    If the limit of the series is not 0, the series diverges, you can simply take limit here and apply L Hoptials rule since the limit is of indeterminate form here, and notice the limit here is not 0, so it diverges, or you can do as acc100jt did, I know when i have a rational fraction with equivalent degree in denominator and numerator the limit is the ratio of the coefficients. However, but are ways of showing how the limit is not 0.
    Last edited by RockHard; December 8th 2009 at 09:37 PM.
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  3. #3
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    a_{n}=\frac{n}{1000n+1}
    and \lim_{n\rightarrow\infty} a_{n}=\lim_{n\rightarrow\infty} \frac{n}{1000n+1}=\lim_{n\rightarrow\infty}\frac{1  }{1000+\frac{1}{n}}=\frac{1}{1000}.

    By Test for Divergence, the series \sum_{n=1}^{\infty}\frac{n}{1000n+1} is divergent
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