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Math Help - converge or diverge problem

  1. #1
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    converge or diverge problem

    Hi! Can someone help me figure out these problems please? I have to determine whether the series converge or diverge and give a reason to the answer. Thanks!

    1.)

    Σ [ln (n+1)]/(n+1)
    n=2


    2.)

    Σ 2/ (1+e^n)
    n=1
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by cherry106
    Hi! Can someone help me figure out these problems please? I have to determine whether the series converge or diverge and give a reason to the answer. Thanks!

    1.)

    Σ [ln (n+1)]/(n+1)
    n=2
    Diverges as \frac{\ln(n+1)}{n+1}>\frac{1}{n+1} for n>e-1 and \sum_{n=2}^\infty \frac{1}{n+1} diverges
    as it is just the tail of the harmonic series.

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by cherry106
    2.)

    Σ 2/ (1+e^n)
    n=1
    Converges, as 1+e^n>n^2 for n>0, so \frac{1}{1+e^n}<\frac{1}{n^2} and \sum_{n=1}^{\infty } \frac{1}{n^2} converges.

    RonL
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  4. #4
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    Thanks for the help!

    For problems like that how do you know what to compare the equation to? for example in question one you compared the denominator to {1}/{n+1} for n>e-1 and for question two you used n^2 for n>0.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by cherry106
    Thanks for the help!

    For problems like that how do you know what to compare the equation to? for example in question one you compared the denominator to {1}/{n+1} for n>e-1 and for question two you used n^2 for n>0.
    In question one we have something slowly increasing over n+1, so the whole
    thing decreases more slowly that 1/(n+1) so we try comparing the series with
    the harmonic series.

    In the second we have essentially 1/(1+e^n), but e^n increases more
    rapidly than any power of n, so 1/(1+e^n) decreases more rapidly than
    the reciprocal of any power of n, and as

    \sum 1/n^2

    is the first of these series which converges its what we use for a comparison
    series.

    RonL
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