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Thread: 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 $\displaystyle x_{n}=\cos{(n\pi/3)}$. Find an increasing list of numbers $\displaystyle 1\leq n_{1}<n_{2}<n_{3}<...$, so that the subsequence $\displaystyle y_{k}=x_{n_{k}}$ tends to 1. Prove that $\displaystyle \{x_{n}\}_{n=1}^{\infty}$ 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 $\displaystyle x_{n}=\cos{(n\pi/3)}$. Find an increasing list of numbers $\displaystyle 1\leq n_{1}<n_{2}<n_{3}<...$, so that the subsequence $\displaystyle y_{k}=x_{n_{k}}$ tends to 1. Prove that $\displaystyle \{x_{n}\}_{n=1}^{\infty}$ does not tend to a limit.
    Consider $\displaystyle n_k = 6\cdot k$. Then $\displaystyle \left( {\forall k} \right)\left[ {\cos \left( {\frac{{n_k \pi }}{3}} \right) = ?} \right]$

    This time,
    Consider $\displaystyle n_j = 9\cdot (2j+1)$. Then $\displaystyle \left( {\forall j} \right)\left[ {\cos \left( {\frac{{n_j \pi }}{3}} \right) = ?} \right]$
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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?

    $\displaystyle n_1=6,~n_2=12,~n_3=18,~...$
    $\displaystyle {\cos \left( {\frac{{n_1 \pi }}{3}} \right) = ?}$
    $\displaystyle {\cos \left( {\frac{{n_2 \pi }}{3}} \right) = ?}$
    $\displaystyle {\cos \left( {\frac{{n_3 \pi }}{3}} \right) = ?}$
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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 $\displaystyle n_{1}=6$ 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 $\displaystyle n_{1}=6$ 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 $\displaystyle x_{n}=\sin{(n\pi/2)}$.Find an increasing list of numbers $\displaystyle 1\leq n_{1}<n_{2}<n_{3}<...$, so that the subsequence $\displaystyle y_{k}=x_{n_{k}}$ tends to 0. Prove that $\displaystyle \{x_{n}\}^{\infty}_{n=1}$ does not tend to a limit.

    The answer for the first bit says...

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

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

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