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Thread: Determine whether a sequence converge

  1. #1
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    Determine whether a sequence converge

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

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

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


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

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

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


    So yes, the sequence converges.
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    $\displaystyle |\sin n|^{\frac{1}{m}}<1$ not necessarily implies that it converges
    Last edited by elim; Aug 27th 2010 at 08:40 PM.
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  4. #4
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    True, I misinterpreted the question, sorry. All I proved was that it was bounded.

    According to Wolfram,

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

    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...

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

    ... and because ...

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

    ... is ...


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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
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    $\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$
    is not obvious. there are integer $\displaystyle n's$ that make $\displaystyle |\sin n|$ very close to $\displaystyle 0$ hence $\displaystyle -\ln |\sin n|$ is very big.
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  8. #8
    Junior Member qspeechc's Avatar
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    Do we have $\displaystyle |\text{sin} n|\leq n \quad \forall n\in\mathbb{N}$ ?
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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 $\displaystyle |\text{sin} n|\leq n \quad \forall n\in\mathbb{N}$ ?
    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
    $\displaystyle \displaystyle \lim_{n \rightarrow \infty} \frac{\ln |\sin n|}{n}= 0$
    is not obvious. there are integer $\displaystyle n's$ that make $\displaystyle |\sin n|$ very close to $\displaystyle 0$ hence $\displaystyle -\ln |\sin n|$ is very big.
    The consideration of elim is quite correct. By definition say that...

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

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

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

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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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 $\displaystyle n^{1/n}\rightarrow 0$ 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 $\displaystyle n^{1/n}\rightarrow 0$ then we are done...
    $\displaystyle \displaystyle n^{1/n}\rightarrow 1$ :P
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  15. #15
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by Defunkt View Post
    $\displaystyle \displaystyle n^{1/n}\rightarrow 1$ :P
    Oops. lol, I must have been smoking something potent...
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