Problem 18 - Essential Calculus

Question

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

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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.

Key Concepts

Convergence of Series

To determine if a series converges, mathematicians use a variety of tests, one of which is the Integral Test. For a series of the form \(\sum_{n=1}^{\infty} a_n\), we use the Integral Test to establish convergence if three conditions are met: the function must be positive, continuous, and decreasing on the interval \([1, \infty)\). If these conditions are met, and the improper integral \(\int_{1}^{\infty} f(x) \, dx\) converges, the series converges as well. However, if even one condition is not satisfied, the Integral Test cannot be applied. In our case,

• The series is positive.
• It is not continuous at all points.
• It is not always decreasing.

Thus, the Integral Test cannot be applied to this series.

Sin Function

The sine function, denoted as \(\sin(x)\), is a periodic function with values ranging between -1 and 1. Its periodicity means that it repeats its values in regular intervals, specifically every \(2\pi\). This behavior affects other mathematical expressions, especially when used in series.For the series \(\left(\frac{\sin n}{n}\right)^2\), the oscillation between 0 and 1 occurs because \(\sin n\) can take any value between -1 and 1. When squared, these values give a range from 0 to 1. This characteristic is crucial in explaining why this series does not simply meet conventional criteria for convergence like monotonicity.

Non-Continuous Function

A function's continuity is an important factor when considering the Integral Test for convergence. A function must be unbroken across the interval to apply the test. In the case of the series \(\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^{2}\), the function is discontinuous at points where \(n\) is a multiple of \(\pi\).These discontinuities occur because \(\sin n = 0\) at multiples of \(\pi\), leading to terms that are undefined. Since these breaks render the series non-continuous on \([1, \infty)\), the Integral Test is invalidated. This demonstrates the importance of checking continuity before applying any convergence test.

Decreasing Function

For a series to satisfy the conditions of the Integral Test, the function must also be decreasing. This means that as \(n\) increases, the function's value must not increase for any part of the interval \([1, \infty)\).The series \(\left(\frac{\sin n}{n}\right)^2\) does not meet this requirement because its values fluctuate. This fluctuation is due to the oscillatory nature of the sine function, making the overall function non-monotonic. Without a clear decreasing pattern, one crucial condition for the Integral Test is not met, preventing us from concluding the convergence of the series based on this test.