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Math Help - Prove divergence

  1. #1
    MHF Contributor alexmahone's Avatar
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    Prove divergence

    Prove that \frac{1}{1}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+... diverges.
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by alexmahone View Post
    Prove that \frac{1}{1}-\frac{1}{2^2}+\frac{1}{3}-\frac{1}{4^2}+... diverges.
    that's alternative series .... use Leibniz test




     \displaystyle \sum _{n=1} ^{\infty} (-1)^n a_n

    if :

     \displaystyle \lim_{n\to \infty} a_n = 0

    and

     a_n \ge a_{n+1}

    than it converges if not than diverges


    (or show that converge/diverge by the definition)
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by yeKciM View Post
    that's alternative series .... use Leibniz test




     \displaystyle \sum _{n=1} ^{\infty} (-1)^n a_n

    if :

     \displaystyle \lim_{n\to \infty} a_n = 0

    and

     a_n \ge a_{n+1}

    than it converges if not than diverges


    (or show that converge/diverge by the definition)
    The alternating series test can be used to prove convergence, not divergence.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by alexmahone View Post
    The alternating series test can be used to prove convergence, not divergence.
    if series do not converge than, series diverge .... just must state that "based on Leibniz test series diverge" (or show by the definition, using partial sums )
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by yeKciM View Post
    if series do not converge than, series diverge .... just must state that "based on Leibniz test series diverge" (or show by the definition, using partial sums )
    If the alternating series test works, the series converges. But if the alternating series test fails, we cannot say anything about the convergence or divergence of the series.
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  6. #6
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by alexmahone View Post
    If the alternating series test works, the series converges. But if the alternating series test fails, we cannot say anything about the convergence or divergence of the series.
    hm... when some series converge means that there is all members of the series except them finite many are in region of some point... if that's not the case than it diverges ... but ... never mind that I'm probably off while translating


    try like this :

     \displaystyle \sum _{n=0} ^{\infty} \frac {1}{2n+1} - \frac {1}{(2n+2)^2 }

    and show that diverges by definition of using some of the test's for convergence (Cauchy, D'Alambert , Raabe or by the integral test )


    D'Alambert


    \displaystyle q= \lim_{n \to \infty } \frac {a_{n+1}}{a_n}  = \left\{\begin{matrix}<br />
q<1 & converges \\ <br />
q>1 & diverges \\ <br />
q=1 & ??? <br />
\end{matrix}\right.


    Raabe :

     \displaystyle q= \lim_{n \to \infty } n(\frac {a_{n}}{a_{n+1} }-1 ) = \left\{\begin{matrix}<br />
q>1 & converges \\ <br />
q<1 & diverges \\ <br />
q=1 & ??? <br />
\end{matrix}\right.
    Last edited by yeKciM; September 9th 2010 at 08:26 AM.
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  7. #7
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    Liebniz's test does not apply here because a_n\ge a_{n+1} is not true.
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  8. #8
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Liebniz's test does not apply here because a_n\ge a_{n+1} is not true.
    yes, my bad ... didn't wrote it like (actually didn't see the series correctly at that time) in the post #6

    Liebniz's test could not be used because this is not the alternative series

     \displaystyle \sum _{n=0} ^{\infty} (-1)^n a_n



    Cauchy and D'alambert's test give value of one so it's inconclusive but using Raabe's or integral test series

     \displaystyle \sum _{n=0} ^{\infty} \frac {1}{2n+1} - \frac {1}{(2n+2)^2 }

    diverge
    Last edited by yeKciM; September 11th 2010 at 09:37 AM.
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