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Math Help - Sequence question

  1. #1
    Senior Member chella182's Avatar
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    Sequence question

    First off, sorry if this is in the wrong section.
    The question goes...

    We define a sequence by x_{n}=\cos{(n\pi/3)}. Find an increasing list of numbers 1\leq n_{1}<n_{2}<n_{3}<..., so that the subsequence y_{k}=x_{n_{k}} tends to 1. Prove that \{x_{n}\}_{n=1}^{\infty} does not tend to a limit.

    I have no clue. I hate stuff like this.
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  2. #2
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    Quote Originally Posted by chella182 View Post
    We define a sequence by x_{n}=\cos{(n\pi/3)}. Find an increasing list of numbers 1\leq n_{1}<n_{2}<n_{3}<..., so that the subsequence y_{k}=x_{n_{k}} tends to 1. Prove that \{x_{n}\}_{n=1}^{\infty} does not tend to a limit.
    Consider n_k = 6\cdot k. Then \left( {\forall k} \right)\left[ {\cos \left( {\frac{{n_k \pi }}{3}} \right) = ?} \right]

    This time,
    Consider n_j = 9\cdot (2j+1). Then \left( {\forall j} \right)\left[ {\cos \left( {\frac{{n_j \pi }}{3}} \right) = ?} \right]
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  3. #3
    Senior Member chella182's Avatar
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    Wow, no idea.
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  4. #4
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    Quote Originally Posted by chella182 View Post
    Wow, no idea.
    What does that mean? Can you do the basic caculations?

    n_1=6,~n_2=12,~n_3=18,~...
    {\cos \left( {\frac{{n_1 \pi }}{3}} \right) = ?}
    {\cos \left( {\frac{{n_2 \pi }}{3}} \right) = ?}
    {\cos \left( {\frac{{n_3 \pi }}{3}} \right) = ?}
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  5. #5
    Senior Member chella182's Avatar
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    Means I have no idea what you're talking about. Where are you getting the n_{1}=6 et cetera from? I just don't understand it at all.
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  6. #6
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    Quote Originally Posted by chella182 View Post
    Means I have no idea what you're talking about. Where are you getting the n_{1}=6 et cetera from? I just don't understand it at all.
    If that is true, then your difficulties extend far beyond any help you can realistically expect to get anywhere on a website. You need a good ole sit-down with a live instructor.
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  7. #7
    Senior Member chella182's Avatar
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    Okay, I have a similar example with a solution, but I don't understand it either.

    We define a sequence by x_{n}=\sin{(n\pi/2)}.Find an increasing list of numbers 1\leq n_{1}<n_{2}<n_{3}<..., so that the subsequence y_{k}=x_{n_{k}} tends to 0. Prove that \{x_{n}\}^{\infty}_{n=1} does not tend to a limit.

    The answer for the first bit says...

    A subsequence where the value is zero is given by n_{k}=2k, so 2<4<6<8...

    Why 2k though?
    The answer to the second bit it says...

    To prove the sequenece does not tend to a limit, we have to find a subsequence where the limit is not zero. Take n_{k}=4k+1, then \sin{((4k+1)\pi/2)}=\sin{(2k\pi+\pi/2)}=1. Now we say, if the sequence \{x_{n}_{n=1}^{\infty} tend to a limit l, then so do the subsequences, hence 0=l=1, which is a contradiction, hence the assumption \{x_{n}\}_{n=1}^{\infty} tends to a limit is false, and the sequence does not tend to a limit.

    Why n_{k}=4k+1? And why does \sin{(2k\pi+\pi/2)}=\sin{(\pi/2)}? I just don't get it.
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