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Posted by on Sunday, July 31, 2011 at 8:12pm.

Show that the series(−1)^n−1(bn) where bn = 1/(n^1/2) if n is odd and bn = 1/2^n
if n is even, diverges. Why does the alternating series test fail?

  • calculus - , Monday, August 1, 2011 at 5:43am

    The series can be considered as the sum of two series S1 & S2:
    for n odd:
    S1 =-∑ 1/sqrt(n)
    for n even:
    S2 = ∑ 1/2^n

    Assuming the summation is for n=1 to ∞, S2 sums to 1 and S1 diverges (by comparison with ∑1/n).

    The alternating series test requires two conditions:
    1. the limit of the sequence approaches 0 as n-> ∞, which is satisfied.
    2. the sequence of bn must decrease monotonically.

    Write out the terms of bn of the alternating series and check if the second condition is satisfied.

  • calculus - , Monday, August 1, 2011 at 8:46am

    im confused. I was taught that if it is an alternating series then you can only use the alternating series test to see if it was diverging or converging. And i did the alternating series test and all the conditions were met for each series separately.

  • calculus - , Monday, August 1, 2011 at 9:15am

    If you separate the terms into two series, the alternating series test does NOT apply because terms in EACH series have the same sign.

    If you combine the two, as given in the original question, it becomes an alternating series, but the test does not apply because the absolute values of the terms are not monotonically decreasing.

    =-1/sqrt(1) + 1/2² -1/sqrt(3) + 1/2^4 - 1/sqrt(5) + 1/2^6
    =-1 + 0.25 - 0.5771 + 0.0625 - 0.4472 + 0.0156

    Since the sequence
    1, 0.25, 0.5771, 0.0625, 0.4472, 0.0156, ...
    is not monotonically decreasing, the alternating series test does not apply, i.e. no conclusion from the test.

    Also, S2 does not sum to 1 as I thought, because it only sums the even terms, i.e.
    S2=∑1/(2n)^2 for n=1,2,....∞
    S1 diverges.
    Therefore the series diverges.

  • calculus - , Monday, August 1, 2011 at 9:28am

    So, you're saying that when you separate an alternating series into two (odd and even) then the (-1)^n-1 goes away? And why did you make S1 negative, but not S2.

    And do i need to find the sum for S2 to answer the question? Because i only know how to find the sum of geometric series and power series, so im not sure how i would approach that.

    Sorry about all the questions, and thank you

  • calculus - , Monday, August 1, 2011 at 10:21am

    If I separate the odd and even terms, all the odd terms are negative, so I factor out the negative sign to make S1 negative (and sum the all positive terms inside the parentheses).

    The part about separating into S1 and S2 is just a technique to demonstrate divergence, and is not part of the answer requested. You do not need to prove divergence since it is stated as part of the question, and you do not need to sum S2 to answer the question.

    The question is simply asking WHY the alternating series test is NOT applicable in this case, namely
    "Why does the alternating series test fail?"

    The expected answer would demonstrate that the given alternating series does not satisfy at least one of the two criteria required for the test to apply. Details of which have been provided in my previous response.

  • calculus - , Monday, August 1, 2011 at 10:47am

    wait. aren't all the odd terms positive? since its (-1)^n-1? because if you sub n=3, it becomes (-1)^2, which is positive?

    And thank you once more. This was very helpful!

  • calculus - , Monday, August 1, 2011 at 11:23am

    Yes, you're right. I had the sign inverted, although the arguments remain valid, fortunately.
    Keep up the good work!

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