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Thread: 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:

    $\displaystyle
    c_n = (-1)^n \frac{(n+1)^n}{n^n}
    $

    $\displaystyle
    d_n = \frac{(n+1)^n}{n^{n + 1}}
    $

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

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

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

    $\displaystyle
    c_n = (-1)^n \frac{(n+1)^n}{n^n}
    $

    $\displaystyle
    d_n = \frac{(n+1)^n}{n^{n + 1}}
    $

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

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

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

    $\displaystyle \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

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

    So it is divergent

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