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Math Help - Infinite Positive Series Convergence

  1. #1
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    Infinite Positive Series Convergence

    does series (n= 1 --> inf) sqrt(n)/(4n + 9) converge or diverge?

    I said that it does converge, and I used the comparison test with b(sub n) = n/(4n + 9) got the limit at (1/4) for b(sub n). But the back of the book said I was wrong, what did I do wrong?

    my steps: 0 < a(sub n) < b(sub n)

    b(sub n) converges as limit approaches infinity, so doesn't that mean a(sub n) converges?

    Thanks for any input. Sorry I did not use latex on this.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    We have:

    \displaystyle\lim_{n \to{+}\infty}{\dfrac{\sqrt{n}}{4n+9}:\dfrac{1}{\sq  rt{n}}}=\dfrac{1}{4}\neq 0

    and

    \displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n^{1/2}}

    is divergent, so the given series is also divergent.

    Fernando Revilla
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  3. #3
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    You may be confusing the sequence converging with the series converging.

    In order that the series \sum_{n=0}^\infty a_n converge to any number, it is necessary (but not sufficient) that the sequence \{a_n\} converge to 0.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    You may be confusing the sequence converging with the series converging.

    In order that the series \sum_{n=0}^\infty a_n converge to any number, it is necessary (but not sufficient) that the sequence \{a_n\} converge to 0.
    Certainly your post responds better than mine to Warrenx's question. I didn't pay attention.

    Fernando Revilla
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