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Math Help - Test the series for convergence or divergence.

  1. #1
    Super Member fardeen_gen's Avatar
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    Test the series for convergence or divergence.


    ∑ (-1)^n n / √(n + 2)
    n=1
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by fardeen_gen View Post

    ∑ (-1)^n n / √(n + 2)
    n=1
    By Leibnitz's Criterion this converges because if we define a_n=\frac{n}{\sqrt{n^3+2}} then a_{n+1}\leqslant{a_n} and \lim_{n\to\infty}a_n=0
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  3. #3
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    Hello, fardeen_gen!

    Test for convergence: . \sum^{\infty}_{n=1}(-1)^n\,\frac{n}{\sqrt{n^3+2}}

    The absolute value of the n^{th} term is: . |a_n| \:=\:\frac{n}{\sqrt{n^3+2}}


    \text{Divide top and bottom by }\sqrt{n^3}\!:\;\;\frac{\dfrac{n}{\sqrt{n^3}}} {\dfrac{\sqrt{n^3+2}}{\sqrt{n^3}}} \;=\;\frac{\dfrac{n}{n^{\frac{3}{2}}}} {\sqrt{\dfrac{n^3+2}{n^3}}} . = \;\frac{\dfrac{1}{n^{\frac{1}{2}}}}  {\sqrt{1 + \dfrac{2}{n^3}}}

    Then: . \lim_{n\to\infty}|a_n| \;=\;\lim_{n\to\infty}\left[ \frac{\dfrac{1}{n^{\frac{1}{2}}}}  {\sqrt{1 + \dfrac{2}{n^3}}}\right] \;=\;\frac{0}{1} \;=\;0


    We have an alternating series whose n^{th} term goes to 0.

    . . Therefore, the series converges.

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