91Ó°ÊÓ

The interval of convergence is the set of all values of \(x\) for which a given series converges. For power series like \(\sum a_n x^n\), this interval is pivotal in understanding where the series behaves well. In the provided series, the radius of convergence derived from the root test is 4. This suggests that the series converges for \(-4 < x < 4\). However, endpoints need careful examination to confirm their convergence behavior. At \(x = -4\), the series changes form and diverges according to the limit comparison test. Meanwhile, at \(x = 4\), the series conditionally converges when scrutinized with the alternating series test. Therefore, the interval of convergence becomes \([-4, 4)\), where \([-4\) is inclusive of convergence but 4 is not.

The alternating series test is a valuable tool for determining the convergence of series that alternate in sign. It requires that two conditions be met:

• The absolute value of the terms \(a_n\), \(|a_n|\), continuously decrease, meaning \(|a_{n+1}| \leq |a_n|\).
• The limit of the absolute terms as \(n\) approaches infinity is zero, i.e., \(\lim_{n \to \infty} |a_n| = 0\).

In our series, when \(x = 4\), it becomes \((-1)^n / \ln n\). Here:- The sequence \(1/\ln n\) decreases.- The limit as \(n\) goes to infinity for \(1/\ln n\) is zero.Meeting these conditions, the series at \(x = 4\) converges conditionally.

The root test helps determine the absolute convergence of series using the formula \(\lim_{n \to \infty} |a_n|^{1/n}\). If this limit is less than 1, the series converges; if more than 1, it diverges. When it equals 1, the test is inconclusive.
Applying the root test to the series \((-1)^n x^n / (4^n \ln n)\) involved calculating \(|a_n|^{1/n} = |x| / 4 \cdot (\ln n)^{-1/n}\). Since \((\ln n)^{-1/n}\) approaches 1 as \(n\) grows, the main factor becomes \(|x|/4\). For convergence, \(|x|/4 < 1\) must hold true, giving the set \(-4 < x < 4\) for convergence, pinpointing a radius of 4.

The limit comparison test is effective for determining convergence or divergence by comparing with another series known for its behavior. If comparing
two positive term series \(a_n\) and \(b_n\), and \(\lim_{n \to \infty} a_n/b_n \) is a positive finite number, then both series converge or diverge together.
In the exercise, when \(x = -4\), the series simplifies to \((-1)^n / \ln n\). By choosing \(b_n = 1/\ln n\), which is known to diverge, applying the limit comparison shows the same fate for our \(a_n\). This aligns with the test, confirming the series diverges for \(x = -4\), disqualifying it from the interval of convergence.