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Prove that $\sum_{n=1}^{\infty}{\sin(\frac{\pi}{n})\cos({\pi n})}$ is convergent, but not absolutely convergent.

My solution:


Proof of convergence:

Observe that $\cos(n \pi) = (-1)^{n}$

so $\sum_{n=1}^{\infty}{\sin(\frac{\pi}{n})(-1)^{n}}$ but this is convergent*.


Proof of absolute convergence:

$\sum_{n=1}^{\infty}{|\sin(\frac{\pi}{n})(-1)^{n}|} > \sum_{n=2}^{\infty}{\sin(\frac{\pi}{n})} > \sum_{n=2}^{\infty}\frac{1}{n} = \infty$

Using fact that $\sin(x) > \frac{x}{\pi}$ for $x \in [0, \pi/2]$

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Please check my solution and I need suggestion how should proof*. Intuition tell me that it converges, because we have the alternating sign $(-1)^{n}$, so we add and subtract terms for even and odd $n$.

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    $\begingroup$ For * use alternating series test. $\endgroup$
    – rtybase
    Commented Jul 20, 2019 at 9:55
  • $\begingroup$ But we have $\frac{1}{n} > \sin\left(\frac{\pi}{n}\right)$ for $n = 1$. $\endgroup$ Commented Jul 20, 2019 at 10:25
  • $\begingroup$ @ViktorGlombik of course, I checked this gap. $\endgroup$
    – Blabla
    Commented Jul 21, 2019 at 11:09

2 Answers 2

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Hints. For * use alternating series test and the fact that $\sin{x}$ is ascending/increasing on $\left[0,\frac{\pi}{2}\right]$. First of all $$\lim\limits_{n\to\infty}\sin{\frac{\pi}{n}}= \lim\limits_{n\to\infty}\frac{\pi}{n} \cdot \frac{\sin{\frac{\pi}{n}}}{\frac{\pi}{n}} \rightarrow 0$$ because $\lim\limits_{\ x\to 0}\frac{\sin{x}}{x}=1$. And $$ n>m \geq2 \Rightarrow 0<\frac{\pi}{n}<\frac{\pi}{m}\leq \frac{\pi}{2} \Rightarrow 0<\sin{\frac{\pi}{n}}<\sin{\frac{\pi}{m}}<1$$ As a result, for $a_n={\sin(\frac{\pi}{n})(-1)^{n}}$ we have $\lim\limits_{n\to\infty}a_n=0$ and $|a_n|$ is monotonically decreasing.

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The real fact is that $\sin(x) \leq x$ for $x\in [0, \pi/2]$. But, I think it is not hard to find an appropriate $c$ such that $\sin(x) \geq cx$ for $x\in [0, \pi/2]$. (Try to draw a graph and look at it seriously.)

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  • $\begingroup$ I compute $c= 1/\pi$ $\endgroup$
    – Blabla
    Commented Jul 20, 2019 at 9:16
  • $\begingroup$ @MartinInf1n1ty I think that works, and maybe the optimal one is $c = 2/\pi$. $\endgroup$
    – Seewoo Lee
    Commented Jul 20, 2019 at 9:35
  • $\begingroup$ @MartinInf1n1ty. Since $ \lim_{n\to \infty}\frac {\sin \pi/n}{\pi/n}=1,$ we have $\sin (\pi/n)>\frac {1}{2}(\pi/n)$ for all but finitely many $n\in \Bbb N.$ $\endgroup$ Commented Jul 20, 2019 at 10:02

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