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Thread: Alternating Series Test

  1. #1
    No one in Particular VonNemo19's Avatar
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    Alternating Series Test

    Hi.

    I've got a straight forward alternating series test problem:

    Use the A.S.T. to determine whether the series

    $\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt{n^3+1}}{n^5}$

    converges.

    We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$, but when we try to show that

    $\displaystyle a_{n+1}\leq{a_n}$

    I am having trouble with this inequality. Help?
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  2. #2
    MHF Contributor

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    Quote Originally Posted by VonNemo19 View Post
    Use the A.S.T. to determine whether the series $\displaystyle \displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n\sqrt{n^3+1}}{n^5}$ converges.
    We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$, but when we try to show that
    $\displaystyle a_{n+1}\leq{a_n}$
    I am having trouble with this inequality. Help?
    Can you show that $\displaystyle f(x)=\dfrac{\sqrt{x^3+1}}{x^5}$ is a decreasing function?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    An alternative, express:

    $\displaystyle a_n=\sqrt{\dfrac{1}{n^7}+\dfrac{1}{n^{10}}}$

    Inmediately we obtain $\displaystyle a_{n+1}\leq a_n$ .
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  4. #4
    Junior Member
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    Quote Originally Posted by VonNemo19 View Post
    We know that $\displaystyle \displaystyle\lim_{n\to\infty}a_n=\lim_{n\to\infty }\frac{\sqrt{n^3+1}}{n^5}=0$
    The limit includes $\displaystyle (-1)^n,$ and it tends to zero anyway since boundedness of $\displaystyle (-1)^n.$

    You could also show that the series converges absolutely.
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