Question

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

Short answer

The reason the Integral Test cannot be applied to the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\) is because the function \(f(n) = \frac{(-1)^{n}}{n}\) is not positive on the interval \([1, \infty)\). The Integral Test requires that the function \(f(n)\) be positive, continuous, and decreasing on this interval. In this case, the function \(f(n)\) fails the positivity condition, because it alternates between positive and negative values for different \(n\).

Step-by-step solution

Identify the sequence term function

Looking at the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\), it can be determined that the sequence term function \(f(n)\) is \(\frac{(-1)^{n}}{n}\).

Test for Continuity

The function \(f(n)\) is continuous for all \(n > 0\), except for \(n = 0\), since the function is not defined at \(n = 0\). But since our region of interest is \(n \geq 1\), we can ignore it. Thus, the function \(f(n)\) is continuous on \([1, \infty)\).

Test for positivity

The function \(f(n)\) is not positive for all \(n \geq 1\). Instead, it alternates between positive and negative values, because of the presence of \((-1)^n\) term. Thus, the function \(f(n)\) is not positive on the interval \([1, \infty)\).

Test for monotonic decreasing

Although the magnitude of terms in the sequence |\(f(n)\)| = \(\frac{1}{n}\) is decreasing, \(f(n)\) itself does not consistently decrease because it alternates in sign.