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Math Help - Series/limit problem

  1. #1
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    Series/limit problem

    The problem I had was whether this series converges or diverges.

    \sum\limits_{n=10}^n arcsin\frac{\pi}{\sqrt{n}}

    What I did was say that,

    arcsin\frac{\pi}{\sqrt{n}}\geq\frac{1}{\sqrt{n}} for every n

    and since \sum\limits_{n=10}^n\frac{1}{\sqrt{n}} diverges, so must \sum\limits_{n=10}^narcsin\frac{\pi}{\sqrt{n}}

    The problem here is that, I do not know how to prove that

    arcsin\frac{\pi}{\sqrt{n}}\geq\frac{1}{\sqrt{n}}

    although it seems plausible.
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  2. #2
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    Re: Series/limit problem

    I did figure out how to solve this with the comparison test from reading older posts. But can it be done the way I described?
    Last edited by HK47; May 8th 2013 at 06:21 AM.
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  3. #3
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    Re: Series/limit problem

    If you can prove that \sin{x}\le{x} for x\ge{0}, then you can take the arcsin of both sides and replace x with \frac{1}{\sqrt{x}}.

    To prove that \sin{x}\le{x} for x\ge{0}, first note that it is obvious for x\ge{1}. For x<1, use the mean value theorem for \sin{x} on [0,x]: \frac{\sin{x} - \sin{0}}{x - 0} = \cos{c} for some c in the interval [0, x], so \frac{\sin{x}}{x}\ge\cos{c}\ge{1}.

    - Hollywood
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  4. #4
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    Re: Series/limit problem

    I'm not sure about your use of the mean value theorem. Doesn't it just say, that there is a point, in that range, where the derivative is equal to the mean change of the functions value, not that the derivative is less than the mean change of value in the whole range, which would need to be proven. sinx\leq x does get proven in the  lim_{x\to 0}\frac{sinx}{x}however.
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  5. #5
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    Re: Series/limit problem

    Oh, i just thought of something neat! Since arcsin(x) is the inverse function of sinx, then if the sinx\leq x then the inverse function reflects sinx around y=x , which would make arcsinx\geq x
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  6. #6
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    Re: Series/limit problem

    Quote Originally Posted by HK47 View Post
    I'm not sure about your use of the mean value theorem. Doesn't it just say, that there is a point, in that range, where the derivative is equal to the mean change of the functions value, not that the derivative is less than the mean change of value in the whole range, which would need to be proven. sinx\leq x does get proven in the  lim_{x\to 0}\frac{sinx}{x}however.
    That's exactly correct. At some point c between 0 and x, the derivative is equal to the mean change. But I'm using that to prove something about the mean change, not about the derivative. Since the derivative is less than or equal to 1, the mean change is less than or equal to 1. And since x was an arbitrary number between 0 and 1, the mean change is less than or equal to 1 for all x between 0 and 1.

    I should have put \frac{\sin{x}}{x} = \cos {c} \le 1, by the way. Sorry about that.

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  7. #7
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    Re: Series/limit problem

    Thanks for the last post. It really made me think .. alot. Anyway, I think the reasoning is correct, but it is not a proof, you use the derivative of sinx, which is proven based on the same sinx is smaller than x, that you set out to prove
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  8. #8
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    Re: Series/limit problem

    No, the original problem was to compute the sum \sum_{n=10}^\infty \sin^{-1}\frac{\pi}{\sqrt{n}}, and \frac{d}{dx}\sin{x}=\cos{x} should be proven by the time you're working on infinite series.

    - Hollywood
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  9. #9
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    Re: Series/limit problem

    Sorry, my last post was unclear. I am saying that in order to prove that (sinx)'=cosx you must first prove limit_{x\to\0}\frac{sin(x)}{x}. And to prove limit_{x\to\0}\frac{sin(x)}{x} you must first prove. x\leq{sin(x)}. So using (sinx)'=cosx to prove x\leq{sin(x)} doesnt function as a proof, as the proof is implicit in the derivative.
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  10. #10
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    Re: Series/limit problem

    Once you've proved that the derivative of \sin{x} is \cos{x}, you're free to use it to prove other stuff.

    You use \lim_{x\to{0}}\frac{\sin{x}}{x}=1 to prove that the derivative of \sin{x} is \cos{x}, but the limit is derived from geometric arguments. The fact that x\le\sin{x} is not involved in the process.

    - Hollywood
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  11. #11
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    Re: Series/limit problem

    The geometric argument proves that sinx\leq{x}\leq{tanx}. That is the purpose of the geometric argument.
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  12. #12
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    Re: Series/limit problem

    Ok, you're right. So we've actually already proven \sin{x}\le{x} (let me get the inequality in the right direction this time) for x\ge{0}.

    - Hollywood
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