The conversation between human and AI assistant.
Is [math]\displaystyle\sum_{n=1}^{\infty} 2 \sin\left(\frac{2}{n}\right) [/math] convergent or divergent?
Is there another method to show that it's divergent?
Yes, the direct comparison test can also be used. We can compare the given series with [math]\displaystyle\sum_{n=1}^{\infty} \frac{2}{n}[/math]. Since [math]\frac{2}{n}[/math] is greater than or equal to [math]2\sin{\frac{2}{n}}[/math], the given series is greater than or equal to [math]\displaystyle\sum_{n=1}^{\infty} \frac{2}{n}[/math]. And since [math]\displaystyle\sum_{n=1}^{\infty} \frac{2}{n}[/math] is divergent, the given series is also divergent.