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Math Help - Proving that a series diverges - help!

  1. #1
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    Proving that a series diverges - help!

    So i've been doing some work on sequences lately and got stuck on this question (it has been attached as an image) Proving that a series diverges - help!-divergence.png

    I know that I have to use the nth term test for divergence but I am not sure of how to go about doing it

    any help is much appreciated! thank you
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    Re: Proving that a series diverges - help!

    Quote Originally Posted by kaya2345 View Post
    So i've been doing some work on sequences lately and got stuck on this question (it has been attached as an image) Click image for larger version. 

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    I know that I have to use the nth term test for divergence.
    You should know that {\lim _{n \to \infty }}{\left( {1 + \frac{a}{{n + b}}} \right)^{cn}} = {e^{ac}}

    Well {\lim _{k \to \infty }}{\left( {\frac{k}{{k + 1}}} \right)^k} = {\lim _{k \to \infty }}{\left( {1 + \frac{{ - 1}}{{k + 1}}} \right)^k}
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    Re: Proving that a series diverges - help!

    Alternatively, represent \left(\frac{x}{x+1}\right)^x as e^{x\ln(x/(x+1))} and then prove that x\ln(x/(x+1))\to-1 as x\to\infty using either Taylor series or L'Hopital's rule.
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  4. #4
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    Re: Proving that a series diverges - help!

    Hello, kaya2345!

    We are expected to know this definition:

    . . \lim_{n\to\infty}\left(1+\tfrac{1}{n}\right)^n \;=\;e


    Prove that that the following series diverges: . \sum^{\infty}_{k=1}\left(\frac{k}{k+1}\right)^k

    Theorem: If the nth term does not approach zero, the series diverges.

    We have: . a_n \;=\;\left(\frac{k}{k+1}\right)^k \;=\;\left(\frac{1}{\frac{k+1}{k}}\right)^k \;=\;\left(\frac{1}{1+\frac{1}{k}}\right)^k  \;=\;\frac{1}{(1+\frac{1}{k})^k}

    Hence: . \lim_{k\to\infty}a_n \;=\;\lim_{k\to\infty}\frac{1}{(1+\frac{1}{k})^k} \;=\;\frac{1}{\lim(1+\frac{1}{k})^k} \;=\;\frac{1}{e} \;\ne\;0

    Therefore, the series diverges.
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