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Math Help - Determine whether a sequence converge

  1. #1
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    Determine whether a sequence converge

    Does \{ |\sin n|^{\frac{1}{n}}\} converge?
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  2. #2
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    Remember that

    -1 \leq \sin{n} \leq 1 for all n, that means

    |\sin{n}| \leq 1 for all n.


    I assume since this is a sequence that n only takes on postitive integer values, so that would mean \frac{1}{n} is positive and therefore exponentiating would keep the inequality intact.

    So |\sin{n}|^{\frac{1}{n}} < 1^{\frac{1}{n}}

    |\sin{n}|^{\frac{1}{n}} < 1.


    So yes, the sequence converges.
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  3. #3
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    |\sin n|^{\frac{1}{m}}<1 not necessarily implies that it converges
    Last edited by elim; August 27th 2010 at 08:40 PM.
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    True, I misinterpreted the question, sorry. All I proved was that it was bounded.

    According to Wolfram,

    \lim_{n \to \infty}(|\sin{n}|^{\frac{1}{n}}) = 1

    so yes, the sequence converges.
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  5. #5
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    Thanks a lot Sir!
    what is the url at Wolfram?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Is...

    \displaystyle |\sin n|^{\frac{1}{n}} = e^{\frac{\ln |\sin n|}{n} (1)

    ... and because ...

    \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0 (2)

    ... is ...


    \displaystyle \lim_{n \rightarrow \infty} |\sin n|^{\frac{1}{n}} = 1 (3)

    Kind regards

    \chi \sigma
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  7. #7
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    \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0
    is not obvious. there are integer n's that make |\sin n| very close to 0 hence -\ln |\sin n| is very big.
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  8. #8
    Junior Member qspeechc's Avatar
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    Do we have |\text{sin} n|\leq n \quad \forall n\in\mathbb{N} ?
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    A Plied Mathematician
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    Reply to Elim at Post # 5:

    Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.
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  10. #10
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    Quote Originally Posted by qspeechc View Post
    Do we have |\text{sin} n|\leq n \quad \forall n\in\mathbb{N} ?
    YES
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Reply to Elim at Post # 5:

    Prove It is most likely referring to WolframAlpha. It's basically a way to enter single Mathematica commands. At least, that's what I use it for. Quite handy.
    Thanks a lot Ackbeet. Very nice to know the online tool.
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  12. #12
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by elim View Post
    \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0
    is not obvious. there are integer n's that make |\sin n| very close to 0 hence -\ln |\sin n| is very big.
    The consideration of elim is quite correct. By definition say that...

    \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0 (1)

    ... means that for any \varepsilon >0 it exists an integer N so that \forall n>N is...

    \displaystyle \frac{|\ln |\sin n||}{n}< \varepsilon (2)

    Now if we measure angles in degree [for example] instead of in radians \forall n|180 is \sin n =0 and the sequence \displaystyle a_{n} = |\sin n|^{\frac{1}{n}} has no limit. If the angles are in radians then \sin n never vanishes and in my opinion the (1) is 'probably true', even if for \varepsilon 'very small' the (2) is satisfied for 'enormous' values of N. In any case a rigorous prove has to be supplied and that is not a trivial job ...

    Kind regards

    \chi \sigma
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  13. #13
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by elim View Post
    YES
    I don't mean to be irritating, but if you can show n^{1/n}\rightarrow 0 then we are done...
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  14. #14
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    Quote Originally Posted by qspeechc View Post
    I don't mean to be irritating, but if you can show n^{1/n}\rightarrow 0 then we are done...
    \displaystyle n^{1/n}\rightarrow 1 :P
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  15. #15
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by Defunkt View Post
    \displaystyle n^{1/n}\rightarrow 1 :P
    Oops. lol, I must have been smoking something potent...
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