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Math Help - Series of positive and negative terms

  1. #1
    Junior Member simplysparklers's Avatar
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    Series of positive and negative terms

    Hey guys!
    Final question....finally! Can anyone help me with it please??

    Investigate the convergence of the series
    \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}, \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}

    Thanks guys!
    Jo
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Hey guys!
    Final question....finally! Can anyone help me with it please??

    Investigate the convergence of the series
    \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}, \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}

    Thanks guys!
    Jo
    The alternating series test is your good friend for testing both series ....


    p.s. They both converge (the alternating series test is my good friend too).
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  3. #3
    Super Member PaulRS's Avatar
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    Ok, note that a_n=\frac{1}{2n-1} is a decreasing sequence, and \lim a_n=0, thus by Leibniz Test \sum{a_n\cdot{(-1)^{n-1}}} converges

    Alternating series test - Wikipedia, the free encyclopedia

    You can do the same for the second one, or see that \sum{\frac{1}{n^{3/2}}} converges, thus \sum{\frac{(-1)^{n-1}}{n^{3/2}}} must be convergent
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by PaulRS View Post
    Ok, note that a_n=\frac{1}{2n-1} is a decreasing sequence, and \lim a_n=0, thus by Leibniz Test \sum{a_n\cdot{(-1)^{n-1}}} converges

    Alternating series test - Wikipedia, the free encyclopedia

    You can do the same for the second one, or see that \sum{\frac{1}{n^{3/2}}} converges, thus \sum{\frac{(-1)^{n-1}}{n^{3/2}}} must be convergent
    Just to clarify on PaulRS's completely correct statement

    If \sum{|a_n|} converges this means that it converges absolutely

    and if a series converges absolutely it converges conditionally(with the (-1)^n


    But the converse is not necassarily true

    I will use these two as examples

    For the first one we first test for absolute convergence

    So we check if \sum_{n=1}^{\infty}\bigg|\frac{(-1)^{n-1}}{2n-1}\bigg|=\sum_{n=1}^{\infty}\frac{1}{2n-1} converges

    To this we notice that all the terms are positive and a{n+1}<a_n

    So the integral test applies

    So we set up the integral test

    \int_1^{\infty}\frac{dn}{2n-1}=\frac{1}{2}\ln|2n-1|\bigg|_1^{\infty}=\infty

    and so since the integral diverges the series diverges. So we know that this series is not absolutely convergent

    but now we check for conditional convergence. To do this we use the alternating series test which states that if a series of the form \sum(-1)^na_n and \exists{N}\backepsilon\forall{n>N}\text{ }a_{n+1}<a_n which means that if you can pick any number \in\mathbb{N} and after than number a_n is monotonically decreasing ( a_{n+1}<a_n)

    If this applies then we can use the altnerating series test which states

    that if \lim_{n\to\infty}a_n=0 the series is convergent

    if \lim_{n\to\infty}a_n\ne{0} the series is divergent


    So since our series \sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}

    Meets this criteria (e.g. \forall{n}>0\text{ }a_{n+1}<a_n

    the alternating series test applies

    So we check to see if \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{2n-1}=0


    since direct substitution yields no indeterminate forms and its value is 0 we conclude the limit is 0

    Therefore this series converges by the alternating series test

    That showed that the converse of "if a series converges absolutely it converges conditionally" is untrue

    --------------------------------------------------------------------------
    Now for your other series

    we check for absolute convergence

    so first we find \sum{|a_n|}=\sum_{n=1}^{\infty}\bigg|\frac{(-1)^{n-1}}{n^{\frac{3}{2}}}\bigg|=\sum_{n=1}^{\infty}\fra  c{1}{n^{\frac{3}{2}}}


    So either seeing that this is a convergent p-series or seeing that

    \int_1^{\infty}\frac{dn}{n^{\frac{3}{2}}}=\frac{-2}{\sqrt{x}}\bigg|_1^{\infty}=2

    so since the integral converges the series converges.

    Now since we have shown that the series absolutely converges it MUST by definition converege conditionally
    -------------------------------------------------------------------------

    Remember to follow instructions though. Generally when they ask you to find the convergence they mean conditional not absolute
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