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Math Help - Converge?

  1. #1
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    Converge? any aolternative methods of explaining?

    I have this issue about some sums:

    \sum_{k=1}^{\infty} \frac{1}{3^k-2^k}
    It must be divergent because 3^k>2^k but I cant show that, any idéas?
    Ohh yeah I tried with 3^k=2^{k*ln(3)} but that didn´t really lead me anywhere,

    And the second one is \sum_{k=1}^{\infty} \frac{3k^2+3k+1}{k^3(k+1)^3}
    I get it to converge to 0 because the denominator goes towards infinity so much faster, but the books answer is 1. Dont know how to get that.
    Any ideas what I am doing wrong?
    Last edited by Henryt999; February 12th 2010 at 05:50 AM.
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  2. #2
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    Krizalid's Avatar
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    huh?

    we can write 3^{k}-2^{k}=3^{k}\left( 1-\frac{2^{k}}{3^{k}} \right), now apply the root test, what do you get?

    as for the second one, we can consider the following bounding: \frac{3k^{2}+3k+1}{k^{3}(k+1)^{3}}<\frac{7k^{2}}{k  ^{6}}=\frac{7}{k^{4}},
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  3. #3
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    well

    Quote Originally Posted by Krizalid View Post
    huh?

    we can write 3^{k}-2^{k}=3^{k}\left( 1-\frac{2^{k}}{3^{k}} \right), now apply the root test, what do you get? Reading about it on wiki

    as for the second one, we can consider the following bounding: \frac{3k^{2}+3k+1}{k^{3}(k+1)^{3}}<\frac{7k^{2}}{k  ^{6}}=\frac{7}{k^{4}},
    dont have a clue how you bounded and got 7k^2 how?
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  4. #4
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    because i'd like to get an upper bound, so the numerator becomes bigger, then obviously for k\ge1 we have that 3k^2+3k+1<3k^2+3k^2+k^2=7k^2.
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  5. #5
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    Sorry but I dont understand

    Quote Originally Posted by Krizalid View Post
    because i'd like to get an upper bound, so the numerator becomes bigger, then obviously for k\ge1 we have that 3k^2+3k+1<3k^2+3k^2+k^2=7k^2.
    So you want an upperbound making the nominator larger, well in this case we ended upp with \frac{7}{k^4}
    The books answer is 1.
    How do I go from \frac{7}{k^4} to 1?

    Also my root test turned up like this, first time so no  \Rightarrow

    3^k-2^k = 3^k(1-\frac{2^k}{3^k})

    \lim_{x_\to \infty} [3^k(1-\frac{2^k}{3^k})]^{1/k}
    <br />
=\lim_{x_\to \infty} 3(1-\frac{2^k}{3^k})^{1/k}
    And here I´m stuck If I multiply by  \frac{1}{3^k} i get [\frac{1}{3^k}-\frac{2^k}{3^{2k}}] so that didn´t work

    But hey still have other algebraic idéas. Lets multiply by \sqrt{3^k} ..and that gives me the original expression to the k:th root multiplied by 3...well....nja that doesnt help....(exept that I understand that if (1-\frac{2^k}{3^k})^{1/k}<\frac{1}{3} then the limit is smaller then 1 and the series converges. Well great thats what I want to find out by still, im back where I started....lets try something else..and here I´m stuck....
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  6. #6
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     3^k - 2^k = 3^k(1 - \frac{2^k}{3^k} ) > \frac{3^k}{2} for all sufficiently large  k .

    Now compare your series to  \sum \frac{2}{3^k}
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