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Math Help - Following series convergent/divergent

  1. #1
    Junior Member utopiaNow's Avatar
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    Following series convergent/divergent

    Here are the 2 series where their nth terms are shown:

    <br />
c_n = (-1)^n \frac{(n+1)^n}{n^n}<br />

    <br />
d_n =  \frac{(n+1)^n}{n^{n + 1}}<br />

    Attempted solution:
    So I know about all the tests: monotone convergence criterion, comparison test, root test, ratio test, and alternating series test.

    For c_n I tried to use the alternating series test however it's neither nonincreasing nor clear if \displaystyle\lim_{n\to\infty}c_n = 0.

    For d_n my hunch is to say that since n \geq 1 \Rightarrow 1^n \leq \frac{(n+1)^n}{n^{n + 1}}. Therefore by the comparison test, since 1^n diverges so does d_n.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by utopiaNow View Post
    Here are the 2 series where their nth terms are shown:

    <br />
c_n = (-1)^n \frac{(n+1)^n}{n^n}<br />

    <br />
d_n = \frac{(n+1)^n}{n^{n + 1}}<br />

    Attempted solution:
    So I know about all the tests: monotone convergence criterion, comparison test, root test, ratio test, and alternating series test.

    For c_n I tried to use the alternating series test however it's neither nonincreasing nor clear if \displaystyle\lim_{n\to\infty}c_n = 0.

    For d_n my hunch is to say that since n \geq 1 \Rightarrow 1^n \leq \frac{(n+1)^n}{n^{n + 1}}. Therefore by the comparison test, since 1^n diverges so does d_n.
    For the first notice that

    \frac{(n+1)^n}{n^n}=\left( \frac{n+1}{n} \right)^n =\left( 1+\frac{1}{n} \right)^n

    This is a very famous limit

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

    So it is divergent

    I hope this helps
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  3. #3
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    d_n=\frac{1}{n}\left(1+\frac{1}{n}\right)^n\longri  ghtarrow{0} as n\to\infty. To see that \sum d_n diverges just notice that \frac{1}{n}<\frac{(n+1)^n}{n^{n+1}}.
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