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Math Help - [SOLVED] Infinite series convergence

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Infinite series convergence

    I think it diverges, but I must find whether or not.
    \sum_{n=1}^{\infty} \sin \big(\frac{1}{n}\big). The problem is that they ask me to use a linear function lesser than the sine function over (0,\frac{\pi}{2}). I have no idea of how to start this.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by arbolis View Post
    I think it diverges, but I must find whether or not.
    \sum_{n=1}^{\infty} \sin \big(\frac{1}{n}\big). The problem is that they ask me to use a linear function lesser than the sine function over (0,\frac{\pi}{2}). I have no idea of how to start this.
    The only linear functions mapping \mathbb{R} into \mathbb{R} are f_a(x)=ax for a\in\mathbb{R} hence the problem boils down to finding a constant a such that \sin x \geq ax for x\in\left(0,\frac{\pi}{2} \right).

    Since (\sin x)''=-\sin x\leq0 for x\in\left[0,\frac{\pi}{2} \right], the sine function is a concave function on this interval thus the line which passes through the points (0,0) and \left( \frac{\pi}{2},1 \right) is under the curve of x\mapsto \sin x. In other words, \boxed{\sin x \geq \frac{2}{\pi}x} as long as x\in\left(0,\frac{\pi}{2} \right).

    If you don't know what a concave function is, you can show that g:x\mapsto \sin x-\frac{2}{\pi}x is non negative on \left[0,\frac{\pi}{2} \right].
    Hint : show that g(0)=g\left(\frac{\pi}{2}\right)=0 and that g is increasing on \left[ 0,\arccos (2/\pi) \right] and decreasing on \left[ \arccos (2/\pi), \frac{\pi}{2}\right]<br />
.

    Note : choosing a=\frac{2}{\pi} is useful (compulsory !) if one uses the fact that x\mapsto \sin x is a concave function. If one uses the second method, a can be any real number which lies in \left( 0, 2/\pi\right].
    Attached Thumbnails Attached Thumbnails [SOLVED] Infinite series convergence-linear_sin.png  
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  3. #3
    MHF Contributor arbolis's Avatar
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    Thanks to both! I know what a concave function is.
    I'm not sure how to finish the exercise. As we have \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}, does that mean that \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n} holds? If yes, then I know how to finish this.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by arbolis View Post
    As we have \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}, does that mean that \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n} holds?
    Yes since if n is a positive integer then \frac{1}{n} \in (0,1] \subset (0,\pi/2).
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by arbolis View Post
    Thanks to both! I know what a concave function is.
    I'm not sure how to finish the exercise. As we have \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}, does that mean that \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n} holds? If yes, then I know how to finish this.
    From flyingsquirels post, you have \sin(1/x) \ge \frac{2}{\pi x} for x \ge 2/\pi.

    RonL
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