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Absolute convergence (2)

Alexmahone

Active member
Jan 26, 2012
268
Using the fact that the interval $\displaystyle\left[2k\pi+\frac{\pi}{4},\ 2k\pi+\frac{3\pi}{4}\right]$ contains an integer, prove that $\displaystyle\sum\frac{\sin n}{n}$ is not absolutely convergent.
 

Krizalid

Active member
Feb 9, 2012
118
$\begin{aligned}
\int_{0}^{n\pi }{\frac{\left| \sin x \right|}{x}\,dx}&=\sum\limits_{j=0}^{n-1}{\int_{j\pi }^{(j+1)\pi }{\frac{\left| \sin x \right|}{x}\,dx}} \\
& =\sum\limits_{j=0}^{n-1}{\int_{0}^{\pi }{\frac{\sin x}{x+j\pi }\,dx}} \\
& \ge \frac{1}{\pi }\sum\limits_{j=0}^{n-1}{\int_{0}^{\pi }{\frac{\sin x}{j+1}\,dx}} \\
& =\frac{2}{\pi }\sum\limits_{j=0}^{n-1}{\frac{1}{j+1}}.
\end{aligned}$

As $n\to\infty$ the magic appears.
 

girdav

Member
Feb 1, 2012
96
Let $n_$ such an integer. What about $\sum_k\frac{|\sin n_k|}{n_k}$?