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Thread: [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.
    $\displaystyle \sum_{n=1}^{\infty} \sin \big(\frac{1}{n}\big)$. The problem is that they ask me to use a linear function lesser than the sine function over $\displaystyle (0,\frac{\pi}{2})$. 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.
    $\displaystyle \sum_{n=1}^{\infty} \sin \big(\frac{1}{n}\big)$. The problem is that they ask me to use a linear function lesser than the sine function over $\displaystyle (0,\frac{\pi}{2})$. I have no idea of how to start this.
    The only linear functions mapping $\displaystyle \mathbb{R}$ into $\displaystyle \mathbb{R}$ are $\displaystyle f_a(x)=ax$ for $\displaystyle a\in\mathbb{R}$ hence the problem boils down to finding a constant $\displaystyle a$ such that $\displaystyle \sin x \geq ax$ for $\displaystyle x\in\left(0,\frac{\pi}{2} \right)$.

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

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

    Note : choosing $\displaystyle a=\frac{2}{\pi}$ is useful (compulsory !) if one uses the fact that $\displaystyle x\mapsto \sin x$ is a concave function. If one uses the second method, $\displaystyle a$ can be any real number which lies in $\displaystyle \left( 0, 2/\pi\right]$.
    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 $\displaystyle \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}$, does that mean that $\displaystyle \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n}$ 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 $\displaystyle \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}$, does that mean that $\displaystyle \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n}$ holds?
    Yes since if $\displaystyle n$ is a positive integer then $\displaystyle \frac{1}{n} \in (0,1] \subset (0,\pi/2)$.
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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 $\displaystyle \sin \big(\frac{1}{x}\big) \geq \frac{2x}{\pi}$, does that mean that $\displaystyle \sin (\frac{1}{n})\geq \frac{2}{\pi} \cdot \frac{1}{n}$ holds? If yes, then I know how to finish this.
    From flyingsquirels post, you have $\displaystyle \sin(1/x) \ge \frac{2}{\pi x}$ for $\displaystyle x \ge 2/\pi$.

    RonL
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