91Ó°ÊÓ

The interval of convergence is a crucial concept when dealing with power series. Here, it defines the range of the independent variable, usually denoted by \( x \), where the series converges. For the series \( \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} \cdot 3^{n}} \), the interval can be determined using the Ratio Test.

By applying the Ratio Test, we find that the series converges as long as \( \left| \frac{x-1}{3} \right| < 1 \). Solving this inequality gets us:

• \( -1 < \frac{x-1}{3} < 1 \)
• Multiplying all terms by 3, we have \( -3 < x-1 < 3 \)
• Adding 1 across the inequality provides \( -2 < x < 4 \)

Thus, the interval of convergence for this series is \((-2, 4)\), indicating the series converges for all \( x \) values between \(-2\) and \(4\), but not including \( -2 \) and \(4\) yet. Checking endpoints will determine if they can be included.

A series is said to converge absolutely if the series of absolute values also converges. This essentially means that rearranging the series does not affect its convergence. It is a stronger form of convergence than conditional convergence.

For the given series, absolute convergence must be confirmed within the interval found from the Ratio Test: \( -2 < x < 4 \). Furthermore, examining whether the series converges at \( x = 4 \) is necessary.Steps to Consider:

• Check within the open interval \( -2 < x < 4 \) to ensure absolute convergence, typically guaranteed by the Ratio Test here.
• At \( x = 4 \), substituting into the series gives \( \sum_{n=1}^{\infty} \frac{1}{n^3} \), which converges as a p-series (\( p=3 > 1\)). Thus, the series converges absolutely at \( x = 4 \).

Hence, the series converges absolutely for any \( x \) within \( -2 < x \leq 4 \).

Conditional convergence occurs when a series converges, but does not converge absolutely. In other words, while the series itself reaches a finite value, the series of its absolute values diverges.

Checking Points for Conditional Convergence:

• The focus is often on endpoints of the interval where absolute convergence was checked.
• For this particular series, check at \( x = -2 \).

When \( x = -2 \), the series transforms into \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \). This is an alternating series, and it converges conditionally, according to the alternating series test.

The alternating series test confirms convergence if the absolute value of each term decreases towards zero. For this case, \,\( \frac{1}{n^3} \) does decrease towards zero as \( n \rightarrow \infty \).

Therefore, the original series exhibits conditional convergence precisely at \( x = -2 \).

The Ratio Test is a fundamental method used to determine whether a series converges or diverges. It involves finding the limit of the absolute ratio of consecutive terms in the series.

How the Ratio Test Works:

• Consider a series \( \sum a_n \).
• Calculate \( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If \( L < 1 \), the series converges absolutely. If \( L > 1 \) or is infinite, the series diverges. If \( L = 1 \), the test is inconclusive.

For the provided series \( \sum_{n=1}^{\infty} \frac{(x-1)^{n}}{n^{3} \cdot 3^{n}} \), we utilize the Ratio Test to determine that
\[ L = \left| \frac{x-1}{3} \right| \]. To achieve convergence, this value must be less than 1, leading to the interval \( -2 < x < 4 \).

The Ratio Test not only helps us find the interval of convergence but also serves as a strong tool for understanding absolute convergence within that interval.