Question

Explain why the Integral Test does not apply to the series. $$ \sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^{2} $$

Short answer

The Integral Test does not apply to the series \( \sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^{2} \) because the series fails to meet two critical conditions of the Integral Test. Firstly, the function is not continuous on the interval \([1, \infty)\) as it becomes undefined for points where \(n\) is a multiple of \(\pi\). Secondly, the terms of the series do not decrease as \(n\) increases.

Step-by-step solution

Check if the function is positive

The function is \(\left(\frac{\sin n}{n}\right)^{2}\), which is always positive because any real number squared is always positive or zero. So the series meets this condition.

Check if the function is continuous

The series seems to be continuous at every point on the interval \([1, \infty)\), except those points where \(n\) is a multiple of \(\pi\). At those points, the value of \(\sin n\) is zero, and the term \(\left(\frac{\sin n}{n}\right)^{2}\) becomes undefined. So, the series is not continuous on the interval \([1, \infty)\). This means the series does not meet this essential criteria outlined by the Integral Test.

Check if the function is decreasing

The series \( \left(\frac{\sin n}{n}\right)^{2}\) does not decrease as \(n\) increases. The values fluctuate between 0 and 1, hence it does not meet the criteria of the Integral Test.