Which of the given series are absolutely convergent?
A) $\sum _ { n = 1 } ^ { \infty } \frac { \sin 2 n } { n }$
B)
$$\sum _ { n = 1 } ^ { \infty } \frac { \cos \frac { \pi n } { 7 } } { n \sqrt { n } }$$

Question

Quizplus · Accepted Answer

To show absolute convergence, we need to show that $\sum_{n=1}^{\infty} \left|\frac{\cos \frac{\pi n}{7}}{n\sqrt{n}}\right|$ converges. We can see that $\frac{|\cos \frac{\pi n}{7}|}{n\sqrt{n}} \leq \frac{1}{n\sqrt{n}}$ for all $n$. Using the $p$-series test with $p=\frac{3}{2}$, we know that $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}$ converges, and therefore, $\sum_{n=1}^{\infty} \left|\frac{\cos \frac{\pi n}{7}}{n\sqrt{n}}\right|$ converges, which implies that $\sum_{n=1}^{\infty} \frac{\cos \frac{\pi n}{7}}{n\sqrt{n}}$ converges absolutely.

On the other hand, $\sum_{n=1}^{\infty} \frac{\sin 2n}{n}$ does not converge absolutely. To see this, note that $\left|\frac{\sin 2n}{n}\right| \geq \frac{\sin^2 n}{n} \geq \frac{1}{2n}$ for all $n \geq 1$. Using the $p$-series test with $p=1$, we know that $\sum_{n=1}^{\infty} \frac{1}{2n}$ diverges, which implies that $\sum_{n=1}^{\infty} \frac{\sin 2n}{n}$ also diverges.

Which of the Given Series Are Absolutely Convergent? A) ∑n=1∞sin⁡2nn\sum _ { n = 1 } ^ { \infty } \frac { \sin 2 n } { n }∑n=1∞​nsin2n​

Which of the Given Series Are Absolutely Convergent?
A) $\sum _ { n = 1 } ^ { \infty } \frac { \sin 2 n } { n }$