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Alternating Series

  1. Dec 9, 2007 #1

    tony873004

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    Science Advisor
    Gold Member

    1. The problem statement, all variables and given/known data
    Test the series for convergence or divergence
    [tex]
    \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^{n - 1} \frac{{\ln n}}{n}}
    [/tex]



    2. Relevant equations
    If [tex]
    b_{n + 1} \le b_n \,\& \,\mathop {\lim }\limits_{n \to \infty } b_n = 0
    [/tex]
    then the series is convergent.



    3. The attempt at a solution

    [tex]
    b_n = \frac{{\ln n}}{n},\,\,\mathop {\lim }\limits_{n \to \infty } b_n = \mathop {\lim }\limits_{n \to \infty } \frac{{\ln n}}{n} = \frac{\infty }{\infty } = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {\ln n} \right)^\prime }}{{\left( n \right)^\prime }} = \mathop {\lim }\limits_{n \to \infty } \frac{{1/n}}{1} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0
    [/tex]
    so it passes the 2nd test.

    [tex]
    \begin{array}{l}
    {\rm{when }}n = 1,\,\frac{{\ln n}}{n} = 0 \\
    {\rm{when }}n = 2,\,\frac{{\ln 2}}{2} = 0.35 \\
    \\
    {\rm{when }}n = 1,\,b_n = 0,\,b_{n + 1} = 0.35\, > 0,\,b_{n + 1} > b_n \\
    \end{array}
    [/tex]
    so it fails the 1st test, hence is divergent

    But the example says:
    [tex]
    \begin{array}{l}
    \left( {\frac{{\ln n}}{n}} \right)^\prime = \frac{{\left( {\ln n} \right)^\prime \left( n \right) - \left( {\ln n} \right)\left( n \right)^\prime }}{{\left( n \right)^2 }} = \frac{{\left( {1/n} \right)n - \left( {\ln n} \right)\left( 1 \right)}}{{n^2 }} = \frac{{1 - \ln n}}{{n^2 }} \\
    \\
    \frac{{1 - \ln x}}{{x^2 }} \le 0\,{\rm{whenever }}x \ge 1 \\
    \end{array}
    [/tex]
    Therefore it passes both tests and is convergent.

    But if I graph the derivative:
    [​IMG]
    I get both positive and negative values past x=1, so is the example wrong when it says that when x>=1 that the derivitave is always <=0? Or did I make a mistake?
     
  2. jcsd
  3. Dec 9, 2007 #2

    quasar987

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    Science Advisor
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    Gold Member

    I guess the example is erroneous about the x>=1 part, but the series is convergent, and your error is that the first point in the criterion should actually be read "if bn is decreasing past some n".

    In books, they usually prove the Leibniz criterion in the form you gave, and also argue somewhere else that a series' nature is independent of the index at which it starts. Combine these two facts to obtain the criterion in the more general form "if bn is decreasing past some n".

    And on a side note: the criterion states "so and so" is a sufficient condition for the series to converge. It does NOT say that "so and so" is a necessary condition, so you cannot conclude that the series diverges just because it fails the test. The classical example is the test "If the series converges, then the general term goes to zero"... this does not means that as soon as the general term of a series goes to zero, the series converges.
     
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