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Math Help - Series question

  1. #1
    Senior Member Pinkk's Avatar
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    Series question

    Evaluate

    \sum_{k=2}^{\infty}\,\,(\frac{2}{3^{k}} + \frac{1}{2^{k}})

    So I separated the two sums and also change the dummy variable using n=k-2 so I can eventually have two geometric series:

    2\sum_{n=0}^{\infty}\,\,(\frac{1}{3})^{n+2} + \sum_{n=0}^{\infty}\,\,(\frac{1}{2})^{n+2}


    Pulling out the constants within the sums, I can use \frac{a}{1-r} for both sums:

    \frac{\frac{2}{9}}{1-\frac{1}{3}} + \frac{\frac{1}{4}}{1-\frac{1}{2}}

    Simplifying, I get:

    \frac{2}{9}\cdot\frac{3}{2}+\frac{1}{4}\cdot\frac{  2}{1}=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}

    However, my book states that the answer is \frac{6}{5}. Did I do something wrong, or is the book's answer key wrong? Any help would be appreciated, thanks.
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  2. #2
    Super Member Gamma's Avatar
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    I got what you got when i just did it myself, but I did it slightly differently. Then looking at yours, yeah i see no flaws. That typo is pretty close, it seems pretty likely that they just messed up.

    You may check online though, a lot of times publishers will put the errata online, it may well mention the mistake there. Just a thought
    Last edited by Gamma; May 1st 2009 at 10:14 PM. Reason: bad grammar
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  3. #3
    Super Member redsoxfan325's Avatar
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    Yes, Maple confirms that the answer is indeed \frac{5}{6}.
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  4. #4
    Senior Member Pinkk's Avatar
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    Thanks guys. I have one more question before I pass out from tiredness:

    \sum_{k=0}^{\infty}\,\,(-1)^{k+1}\frac{3k}{k+1}

    How would one go about solving this. I can "see" that the series will diverge since |a_{N+1}|>|a_{N}| when N \ge 0, essentially as you go farther out the distance a_{k} is from the x-axis increases. How can show that this series diverges without giving such a wordy argument; essentially, how can I evaluate it exactly either by finding an expression for S_{n} or some other method.
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  5. #5
    Super Member Gamma's Avatar
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    It is a necessary condition for a series to converge that the limit as n goes to infinity of the terms go to 0. This is clearly not the case, therefore the series will not convege.
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  6. #6
    Senior Member Pinkk's Avatar
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    Ah, I think I get it, so eventually you'd have it where:

    \lim_{k\to\infty}(-1)^{k+1}\cdot\lim_{k\to\infty}\frac{3k}{k+1}, and since (-1)^{k+1} always oscillates between -1 and 1, \lim_{k\to\infty}a_{k} does not exist and therefore the series diverges.
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  7. #7
    Super Member Gamma's Avatar
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    bingo.

    Another classic one is like \sum_{n=0}^{\infty}sin(n).

    Everyone pulls out all the stops to try to use one of the tests, but forgets to first check the limit of the terms actually is 0. That should always be your first test because it is so easy to do and could really save you a ton of time in the end.
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  8. #8
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    THe answer is 3ln2 - 3k , k is 1 or 0 so the series diverges
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