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Math Help - [SOLVED] simple series question

  1. #1
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    [SOLVED] simple series question

    assume the following series converge: \sum_{j=1}^\infty (d_{2j}+d_{2j+1}) and \sum_{j=1}^\infty (d_{2j+1}+d_{2j+2}), where d_j can have any sign, and the convergence is not necessarily absolute. In general this doesn't imply that d_j\to 0, but after a change of variables we may assume d_j\to 0. Now assuming this, is it true that \sum_{j=1}^\infty d_j is convergent?

    P.S. well, ok, I lied, it's not exactly a "simple question", but it shouldn't be too deep either....
    Last edited by choovuck; June 5th 2010 at 09:19 PM.
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  2. #2
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    Quote Originally Posted by choovuck View Post
    assume the following series converge: \sum_{j=1}^\infty (d_{2j}+d_{2j+1}) and \sum_{j=1}^\infty (d_{2j+1}+d_{2j+2}), where d_j can have any sign, and the convergence is not necessarily absolute. In general this doesn't imply that d_j\to 0, but after a change of variables we may assume d_j\to 0. Now assuming this, is it true that \sum_{j=1}^\infty d_j is convergent?

    P.S. well, ok, I lied, it's not exactly a "simple question", but it shouldn't be too deep either....
    No. d_j=(-1)^j
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    No. d_j=(-1)^j
    no, I'm demanding d_j\to 0

    (basically, if d_{2j}\to c_1 with some c_1\ne0, then d_{2j+1}\to -c_1, and we can make the change of variables \widehat{d_{2j}}=d_{2j}-c_1, \widehat{d_{2j+1}}=d_{2j+1}+c_1, and ask whether \sum_{j=1}^\infty \widehat{d_{j}} is convergent)
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  4. #4
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    I've not been formally trained in series, so I may be making a bad assumption, but if d_j \to 0, doesn't that imply d_{2j} \to 0?
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  5. #5
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    Quote Originally Posted by Turiski View Post
    I've not been formally trained in series, so I may be making a bad assumption, but if d_j \to 0, doesn't that imply d_{2j} \to 0?
    yes, d_j \to 0 implies d_{2j} \to 0 and d_{2j+1} \to 0, but it's in no way helpful
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  6. #6
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    If the sum:
    \sum_{j=1}^\infty (d_{2j}+d_{2j+1})

    is convergent then \lim_{j \rightarrow \infty}{d_{2j}+d_{2j+1}}=0

    same for the second sum \lim_{j \rightarrow \infty}{d_{2j+1}+d_{2j+2}}=0

    Now let k=2j

    \lim_{k \rightarrow \infty}{d_{k}+d_{k+1}}=0 and \lim_{k \rightarrow \infty}{d_{k+1}+d_{k+2}}=0

    Now use the ratio test for absolute convergence for series that need not have positive terms and need not be alternating to use this test.
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  7. #7
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    This is how I approached this:

    Let S = \sum_{j=1}^{\infty} d_{2j} + d_{2j+1}, \ T = \sum_{j=1}^{\infty} d_{2j+1} + d_{2j+2} and  W_n = \sum_{j=1}^{n} d_j.

    Then the partial sums S_n = \sum_{j=1}^{n} d_{2j} + d_{2j+1} and T_n = \sum_{j=1}^{n} d_{2j+1} + d_{2j+2} converge.

    Observe that  \forall n \geq 1, \ S_n = W_{2n+1} - d_1 \ \text{and } \forall n \geq 2, \ T_{n-1} = W_{2n} -d_1 -d_2. Then from the convergence of S_n , \ T_n we get that W_{2n} \text{ and } W_{2n+1} converge (You can actually prove it with only having that W_{2n} converges and d_n \to 0..)

    Now you finish it
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  8. #8
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    Quote Originally Posted by Defunkt View Post
    This is how I approached this:

    Let S = \sum_{j=1}^{\infty} d_{2j} + d_{2j+1}, \ T = \sum_{j=1}^{\infty} d_{2j+1} + d_{2j+2} and  W_n = \sum_{j=1}^{n} d_j.

    Then the partial sums S_n = \sum_{j=1}^{n} d_{2j} + d_{2j+1} and T_n = \sum_{j=1}^{n} d_{2j+1} + d_{2j+2} converge.

    Observe that  \forall n \geq 1, \ S_n = W_{2n+1} - d_1 \ \text{and } \forall n \geq 2, \ T_{n-1} = W_{2n} -d_1 -d_2. Then from the convergence of S_n , \ T_n we get that W_{2n} \text{ and } W_{2n+1} converge (You can actually prove it with only having that W_{2n} converges and d_n \to 0..)

    Now you finish it
    this works, thanks!
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