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Math Help - convergent or divergent

  1. #1
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    convergent or divergent

    classify the following series as absolutely convergent, conditionally convergent or divergent:

    <br />
\sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}

    im not sure what test to use for this one, i tried the Ratio test (d'Alembert) but i didn't really get anywhere! Please help!
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  2. #2
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    Quote Originally Posted by wik_chick88 View Post
    classify the following series as absolutely convergent, conditionally convergent or divergent:
    \sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}
    Would it surprise you to learn that \displaystyle <br />
\sqrt[n]{{\frac{{n!}}<br />
{{n^n }}}} \to \frac{1}<br />
{e}~?
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  3. #3
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    Hello, wik_chick88!

    The Ratio Test applies . . . but it takes a lot of fancy footwork.


    Classify the following series as absolutely convergent, conditionally
    convergent or divergent: . \displaystyle \sum^{\infty}_{n=1} \frac{2.5^{n}n!}{n^n}

    \displaystyle R \;=\;\frac{a_{n+1}}{a_n} \;=\;\frac{2.5^{n+1}(n+1)!}{(n+1)^{n+1}} \cdot\frac{n^n}{2.5^n\,n!} \;=\;\frac{2.5^{n+1}}{2.5^n} \cdot\frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}

    . . \displaystyle =\;2.5(n+1)\cdot\frac{n^n}{(n+1)^{n-1}} \;=\;2.5\cdot\frac{n^n}{(n+1)^n} \;=\;2.5\cdot\left(\frac{n}{n+1}\right)^n


    Divide numerator and denominator by \,n\!:

    . . \displaystyle R \;=\;2.5 \left(\frac{1}{1+\frac{1}{n}}\right)^n \;=\;\frac{2.5}{(1 + \frac{1}{n})^n}


    \displaystyle \text{Take the limit: }\;\lim_{x\to\infty}R \;=\;\lim_{x\to\infty} \frac{2.5}{\left(1+\frac{1}{n}\right)^n} \;=\;\frac{2.5}{\underbrace{\lim\left(1+\tfrac{1}{  n}\right)^n}_{\text{This is }e}}

    Therefore: . \displaystyle \lim_{x\to\infty}R \;=\;\frac{2.5}{e} \;<\;1

    . . The seres converges absolutely.

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  4. #4
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    And if we were all as clever as Plato, the root tests would work nicely!
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