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Math Help - Series Divergence

  1. #1
    Member mohammadfawaz's Avatar
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    Series Divergence

    Hello,

    How do I show that the series \sum_{k=1}^\infty [e-(1+\frac{1}{k})^k] diverges. The question asks to us to use Maclaurin series to compare the series with some other series.

    Thank you
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  2. #2
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    Quote Originally Posted by mohammadfawaz View Post
    The question asks to us to use Maclaurin series to compare the series with some other series.
    So how did it work for you? I guess you wrote \left(1+\frac{1}{k}\right)^k = \exp\left(k\log\left(1+\frac{1}{k}\right)\right) and expanded the exponent first, up to some order, etc...
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  3. #3
    Member mohammadfawaz's Avatar
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    What I actually did is to use the binomial theorem.
    We get an inequality for (1+\frac{1}{k})^k:
    (1+\frac{1}{k})^k<1+1+\frac{1}{2!}+...+\frac{1}{k!  }.
    Expanding e, and taking the difference, I got:
    e-(1+\frac{1}{k})^k>\frac{1}{(k+1)!}+\frac{1}{(k+2)!  }+...
    But I'm stuck there. I don't know if I'm on the right track.

    Thank you
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  4. #4
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    Quote Originally Posted by mohammadfawaz View Post
    e-(1+\frac{1}{k})^k>\frac{1}{(k+1)!}+\frac{1}{(k+2)!  }+...
    Good idea, but the right hand side will give you a convergent series, hence you get nothing. (Greater than a convergent series proves nothing)

    What you could do is use \log(1+x)\geq x-\frac{x^2}{2} for x\geq 0 in the expansion I wrote. Or the asymptotic expansion \log(1+x)=x-\frac{x^2}{2}+o(x^2) as x\to 0.
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