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So I want to decide if this series converges or diverges

$\sum_{n=1}^\infty \sqrt {1- \cos(\pi /n)}$.

My initial thought is that I should calculate

$\lim_{n\to\infty} \sqrt {1- \cos(\pi /n)}$

which approaches zero because $\sqrt{1-1=0}$ and then it converges.

I don't really know if it's correct though as I feel like my operations are a bit weak. What can I add to the solution and am I even right?

Sebastiano
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PythonDaniel
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    If the series converges then the nth term goes to zero. not the other way around. – Cousin Sep 15 '20 at 19:44
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    Just knowing that $\lim a_n=0$ does not tell you that $\sum a_n $ converges (consider $a_n=\frac 1n$ for example). In this case, I'd suggest using the approximation $\cos x\approx 1-\frac {x^2}2$ (for small $x$) for comparison purposes. – lulu Sep 15 '20 at 19:44

2 Answers2

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HINT:

$$\sqrt{1-\cos(\pi/n)}=\sqrt {2}\sin(\pi/(2n))$$

and $\sin(x)\ge \frac2\pi x$ for $0\le x\le\pi/2$.

Mark Viola
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  • So if I use \sqrt {2} as a constant and apply the Series Limit Comparison Test on it, I can see that it diverges? I am not exactly sure why you used the 0≤≤/2 and nothing else, could you please explain that choice to me, I mean how do you know exactly what values to use? :) – PythonDaniel Sep 15 '20 at 20:03
  • Set $x=\pi/(2n)$. Now compare this series to the divergent harmonic series$\sum_{n=1}^\infty \frac1n$. Do you see now how easy this makes things? – Mark Viola Sep 15 '20 at 20:12
  • Yes I got it now I think, by comparing it to a series that are constant/xn we will get the same results as n continuous to grow, right? – PythonDaniel Sep 15 '20 at 20:29
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    Its terms dominate those of the harmonic series. The harmonic series diverges. And so does any dominant series. – Mark Viola Sep 15 '20 at 20:35
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We have that by standard limit

$$\sqrt {\frac{1- \cos\left(\pi /n\right)}{(\pi /n)^2}} \to \frac1{\sqrt 2}$$

then refer to limit comparison test.

user
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