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Alternating Series Convergence Test

  1. Jun 12, 2013 #1
    According to my calculus book two parts to testing an alternating series for convergence. Let s = Ʃ(-1)n bn. The first is that bn + 1 < bn. The second is that the limn[itex]\rightarrow[/itex]∞ bn = 0. However, isn't the first condition unnecessary since bn must be decreasing if the limit is zero. I can't think of any example where the limit will be zero and the function is increasing (assuming bn is not negative which would not really make sense since the series is alternating).
  2. jcsd
  3. Jun 12, 2013 #2
    What if ##b_{2n} = 0## and ##b_{2n+1} = \frac{1}{2n+1}##. That will give a divergent series.
  4. Jun 12, 2013 #3


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    [tex]\frac{sin^{2}(n)}{n}[/tex] is not monotone decreasing.
  5. Jun 12, 2013 #4
    Yes but although sin2(x)/x is not always decreasing the series still converges. I just don't really see the point of the first test. Especially since this series seems to converge and is at times increasing.
  6. Jun 12, 2013 #5
    A test will never work for all series. For any convergence test, there will be a convergent series that is not described by the test.

    The alternating series test is handy because it says, for example, that

    [tex]\sum \frac{(-1)^n}{n}[/tex]

    converges. Furthermore, it gives a handy approximation that if ##\sum (-1)^n b_n## is your series, then it converges to x and

    [tex]\left|x - \sum_{n=1}^k (-1) b_n\right|\leq b_k[/tex]

    So it tells you how close you are to your limit.
  7. Jun 12, 2013 #6


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    You asked for a sequence which is not strictly monotone decreasing but still satisfies ##\lim_{n\rightarrow \infty}a_n = 0##. The above is such an example. Note that not strictly monotone decreasing doesn't imply strictly monotone increasing. A sequence can be neither monotone increasing nor decreasing.

  8. Jun 12, 2013 #7
    I understand this, but what I am wondering is why my book says that each successive term must be smaller than the last. Shouldn't all alternating series in which bn goes to 0 as n approaches infinity be convergent? If this is true then what is the point of the first condition?
  9. Jun 12, 2013 #8
    See post 2 for a counterexample.
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