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Let $\alpha > 0$.

$$\forall n \in \mathbb{N}^*, u_n = 1- \cos{\frac{1}{n^\alpha}}$$

How do I know when the series $\sum u_n$ converges depending on the value of $\alpha$ ?

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1  
Taylor expand maybe? –  mixedmath Sep 16 '12 at 15:09

2 Answers 2

up vote 1 down vote accepted

Write: $$ u_n = 1 - \cos\left(\frac{1}{n^\alpha}\right) = 2 \sin^2\left(\frac{1}{2 n^\alpha}\right) < \frac{1}{2 n^{2\alpha}} $$ Moreover $\lim_{n\to\infty} 2 n^{2 \alpha} u_n = 1$. Thus the series $\sum_{n=1}^{\infty} u_n$ converges when $\sum_{n=1}^\infty n^{-2 \alpha}$ does.

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Whatever $\alpha>0$ is, $n^{\alpha} \to \infty$ so $\cos (1/n^{\alpha}) $ will go to $\cos 0 =1$ so $u_n \to 0.$


By Taylor expanding, we get $1- \cos(1/n^{\alpha}) = 1/( 2 n^{2\alpha} ) + O( 1/n^{4\alpha}).$ So the sum converges if and only if $\alpha >0.5$.

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3  
But the word used was serie(s), not sequence, so I would guess the question is really about $\sum u_n$. –  Harald Hanche-Olsen Sep 16 '12 at 15:11
    
Right, I was talking about $\sum u_n$ –  Skydreamer Sep 16 '12 at 15:14

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