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Math Help - comparison test

  1. #1
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    comparison test

    sum (n+1)/(n+2) as n goes from 1 to infinity

    my solution:
    let {Xn}= (n+1) /(n+2)
    for large n , the series is approx. to 1
    where sum 1 is divergent.
    so i am going to show that {Xn} >= C .1 where C is a constant
    in fact, n+1>n and n+2=< 3n
    so Xn=(n+1)/(n=2)> n/3n=(1/3) . 1
    since sum 1 diverges, so does sum (1/3).1
    so by comparison test, sum Xn is divergent.

    is my proof correct? anybody help me? many thanks~!
    Last edited by xixihaha; April 26th 2009 at 08:09 AM.
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  2. #2
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    This series fails the first test: \frac{{n + 1}}{{n + 2}} \to 1 \ne 0.
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  3. #3
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    thanks. i know that is the simplest way to do.
    but now i am asked to use the comparison test, is my solution correct? Thanks
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  4. #4
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    \left( {\forall n} \right)\left[ {\frac{{n + 1}}<br />
{{n + 2}} > \frac{1}<br />
{2}} \right] so \sum\limits_{n = 1}^K {\frac{{n + 1}}<br />
{{n + 2}}}  > \frac{K}{2}.
    Make K as large as needed.
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