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 $\alpha$ ?
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 $\alpha$ ?
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.
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 $\alpha >0.5$.