Problem 30 - Thomas Calculus

Question

In Exercises \(1-36,\) (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely (c) conditionally? $$ \sum_{n=2}^{\infty} \frac{x^{n}}{n \ln n} $$

Step-by-Step Solution

Answer

The series converges absolutely on \((-1, 1)\), conditionally at \(x = -1\), and diverges at \(x = 1\).

Step 1: Identify the General Term of the Series

The series given is \( \sum_{n=2}^{\infty} \frac{x^{n}}{n \ln n} \). We can identify the general term as \( a_n = \frac{x^n}{n \ln n} \).

Step 2: Use the Ratio Test for Convergence

To find the radius of convergence, apply the Ratio Test. Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This gives us \[ \lim_{n \to \infty} \left| \frac{x^{n+1} (n \ln n)}{x^n ((n+1) \ln(n+1))} \right| = \lim_{n \to \infty} \left| x \right| \cdot \frac{n \ln n}{(n+1) \ln(n+1)}. \] Simplifying, we find \( \lim_{n \to \infty} \left| x \right| \cdot 1 = \left| x \right|. \)

Step 3: Determine the Radius of Convergence

For the Ratio Test to conclude convergence, \( \left| x \right| < 1 \) is required. Therefore, the radius of convergence is \( R = 1 \).

Step 4: Find the Interval of Convergence

Start with \( \left| x \right| < 1 \), which corresponds to the interval \( (-1, 1) \). Now test the endpoints, \( x = -1 \) and \( x = 1 \), for convergence.

Step 5: Test Convergence at Endpoint \( x = 1 \)

Substitute \( x = 1 \) into the series: \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \). This is a version of the Harmonic series and diverges by comparison to the Harmonic series \( \sum_{n=2}^{\infty} \frac{1}{n} \).

Step 6: Test Convergence at Endpoint \( x = -1 \)

Substitute \( x = -1 \) into the series: \( \sum_{n=2}^{\infty} \frac{(-1)^n}{n \ln n} \). This series is an alternating series and we apply the Alternating Series Test. Since \( \frac{1}{n \ln n} \) is positive, decreasing, and tends to zero as \( n \to \infty \), the series converges.

Step 7: Series Converges Absolutely Within the Radius

Within the interval \( (-1, 1) \) (not including endpoints), the series converges absolutely. This is because if \( \left| x \right| < 1 \), the absolute series \( \sum_{n=2}^{\infty} \frac{|x|^n}{n \ln n} \) converges by Ratio Test.

Step 8: Series Converges Conditionally at \( x = -1 \)

At \( x=-1 \), as determined before, the series converges, but not absolutely. So it converges conditionally at this point.

Key Concepts

Ratio Test

The Ratio Test is a powerful method in calculus for determining the convergence of a series. It is particularly useful for series with terms involving expressions like powers or factorials.

To apply the Ratio Test, you start by identifying the general term of a series, which is often represented as \(a_n\). You then look at the limit of the absolute value of the ratio of consecutive terms, \(a_{n+1}/a_n\). Specifically, calculate \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.\]

If the result of this limit is less than 1, the series converges absolutely. If it's greater than 1, the series diverges. If the result equals 1, the test is inconclusive and another method must be used to determine convergence. In our case, the Ratio Test revealed a radius of convergence \(R = 1\), meaning \( -1 < x < 1\) is the interval where the series may converge.

Alternating Series Test

The Alternating Series Test helps determine the convergence of series where the terms alternate in sign (i.e., negative, positive, negative, and so forth). This test is particularly handy when you suspect a series converges due to its alternating nature.

To use this test, consider a series with terms \( (-1)^n b_n \), where \( b_n \) are positive terms. The test has three conditions:

• The terms \( b_n \) must be positive.
• The sequence \( b_n \) must be decreasing.
• \( b_n \) approaches 0 as \( n \to \infty \).

For our series at \( x = -1 \), the terms alternate and the sequence \( 1/(n \ln n) \) is positive, decreasing, and tends to zero. Thus, by the Alternating Series Test, the series converges at \( x = -1 \).

Absolute Convergence

Absolute convergence is an important concept when evaluating the overall behavior of series, especially regarding the robust nature of convergence.

A series \( \sum a_n \) converges absolutely if the series of absolute values \( \sum |a_n| \) also converges. This implies that if a series converges absolutely, it also converges regardless of the order of the terms.

Within our established interval \((-1, 1)\), the series converges absolutely if \( \left| x \right| < 1 \). The Ratio Test confirms this because the series of absolute values \(\sum_{n=2}^{\infty} \frac{|x|^n}{n \ln n}\) converges, owing to the limit \(|x| \cdot 1 \). No need for further checks—absolute convergence is guaranteed if \( x \) falls strictly within this interval.

Conditional Convergence

Conditional convergence is when a series converges overall, but not absolutely, meaning the series converges only because of the arrangement or alternation of terms.

In our exercise, we have determined that at \( x = -1 \), the series converges due to its alternating nature, as confirmed by the Alternating Series Test. However, if we take the absolute values, \( \sum_{n=2}^{\infty} \frac{1}{n \ln n} \), the series does not converge due to its resemblance to a divergent harmonic series.

Thus, at \( x = -1 \), the series converges conditionally. This means the convergence is dependent on the terms maintaining an alternating pattern without which the series would diverge. Such conditional convergence highlights the delicate balance needed for this series to sum to a finite limit only in a specific arrangement of its terms.