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Math Help - how to solve this problem?

  1. #1
    Newbie
    Joined
    Nov 2008
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    3

    how to solve this problem?

    i have been thinking for a long time , please help me
    how to solve this problem?-1.bmp
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  2. #2
    Member
    Joined
    Dec 2008
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    228
    1. I believe it's zero because e-e=0....

    There's a segment of the series that equals e.

    I don't know about 2. though.
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  3. #3
    Senior Member
    Joined
    Dec 2007
    From
    Anchorage, AK
    Posts
    276
    I'm prette sure the first sum diverges. One can check that
    e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k} for integer k≥3.
    Now, the harmonic series
    \sum_{n=1}^{\infty}\frac{1}{n} diverges, and so
    \sum_{n=3}^{\infty}\frac{1}{n} diverges as well.
    but since e-\left(1+\frac{1}{k}\right)^k\gt\frac{1}{k} for k≥3
    then
    \sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]\gt\sum_{n=3}^{\infty}\frac{1}{k}
    and so by the comparison test, both sums diverge, and since
    \sum_{n=1}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right]=(e-2)+(e-\frac{9}{4})+\sum_{n=3}^{\infty}\left[e-\left(1+\frac{1}{k}\right)^k\right],
    your series in problem 1 diverges.

    --Kevin C.
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