Problem 29 - 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? $$ \sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}} $$

Step-by-Step Solution

Answer

The radius is 1, with absolute convergence for \(-1 < x < 1\) and conditional convergence at \(x = -1\).

Step 1: Find the Radius of Convergence

To find the radius of convergence, we use the root test or ratio test. We take \[an = \frac{x^{n}}{n(\ln n)^{2}}\]and apply the ratio test: \[\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_{n}}\right| = \lim_{n \to \infty} \left|\frac{x^{n+1}}{(n+1)(\ln (n+1))^{2}} \cdot \frac{n(\ln n)^{2}}{x^{n}}\right| = \lim_{n \to \infty} \left|x \cdot \frac{n (\ln n)^{2}}{(n+1)(\ln(n+1))^{2}}\right|.\]Using limits, we find\[\lim_{n \to \infty} \frac{n}{n+1} = 1\text{ and }\lim_{n \to \infty} \left(\frac{\ln n}{\ln(n+1)}\right)^{2} = 1.\]Thus, \[\lim_{n \to \infty} \left|x\right| = |x| < 1 \text{ for convergence.}\]Therefore, the radius of convergence is 1.

Step 2: Determine the Interval of Convergence

From the previous step, the series converges when \(|x| < 1\). Thus, the interval of convergence is initially \((-1, 1)\). Now, check the endpoints: **For \(x = 1\):**The series becomes \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}\), which diverges by the integral test comparing it to \(\frac{1}{n \ln n}\), known to diverge.**For \(x = -1\):**The series becomes \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n(\ln n)^{2}}\). Again, \(\sum \frac{1}{n(\ln n)^{2}}\) diverges, thus the alternating series \(\sum \frac{(-1)^n}{n(\ln n)^{2}}\) actually converges conditionally by the alternating series test.So, the interval of convergence is \((-1, 1]\).

Step 3: Find Absolute and Conditional Convergence

**Absolutely convergent:** occurs when \(|x| < 1\) as within this interval, the series \(\sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}}\) converges absolutely by the root test.**Conditionally convergent:** is only at \(x = -1\). For this value, the series converges by the alternating series test, but the series \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}\) diverges, indicating only conditional convergence here.

Key Concepts

Radius of Convergence

The radius of convergence is a key concept when dealing with power series. It tells you how far you can go from the center of the series and still have the series converge. To find this, we often use the ratio test, which provides a convenient measure. In the case of the series \( \sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}} \), the ratio test is applied by comparing \(a_{n+1}\) with \(a_{n}\) and taking the limit as \(n\) approaches infinity.

• Starting with \(a_n = \frac{x^{n}}{n(\ln n)^{2}}\), the limit for the ratio test is calculated as \(\lim_{n \to \infty} \left|x \cdot \frac{n(\ln n)^{2}}{(n+1)(\ln(n+1))^{2}}\right|\).
• Here, the terms \(\frac{n}{n+1}\) and \(\left(\frac{\ln n}{\ln(n+1)}\right)^{2}\) both simplify to unity as \(n\) becomes very large.

Therefore, the radius of convergence is 1, meaning the series converges when \(|x| < 1\). This provides a circle of convergence centered on zero with a radius of one.

Interval of Convergence

The interval of convergence is the range of \(x\) values for which our series converges. For the series \( \sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}} \), after determining the radius of convergence to be 1, we examine \(|x| < 1\), suggesting an initial interval of \((-1, 1)\). However, we must also test the endpoints separately, as behaviors at these points may differ.

• At \(x = 1\), the series becomes \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}\), which diverges. This is verified using the integral test by comparing it with \(\frac{1}{n\ln n}\). The latter is well-known to diverge.
• At \(x = -1\), the series changes to \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n(\ln n)^{2}}\). Although the underlying series \(\sum \frac{1}{n(\ln n)^{2}}\) diverges, the alternating series test shows that this specific alternating series converges.

Thus, the final interval of convergence for this series is \((-1, 1]\), indicating understandings at each endpoint are crucial.

Absolute Convergence

Absolute convergence in a series implies that if you take the absolute values of all terms in the series, it still converges. This is a strong form of convergence because it means that the series is well-behaved in terms of its growth.For our series \( \sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}} \), absolute convergence occurs when \(|x| < 1\). Within this range, the terms decrease rapidly enough that their sum remains finite. The series absolutely converges because the root or ratio test confirms convergence given that the terms are bounded by a function that converges.This implies that not only does the series converge, but so does the series of absolute values of its terms, ensuring stability across multiple operations and rearrangements.

Conditional Convergence

Conditional convergence occurs when a series converges, but it does not converge absolutely. Specifically, this means that while the full series sums to a finite number, taking the absolute values of each term would result in an infinite sum.For our specific series \( \sum_{n=2}^{\infty} \frac{x^{n}}{n(\ln n)^{2}} \), conditional convergence is noted at \(x = -1\). Here, the series becomes \(\sum_{n=2}^{\infty} \frac{(-1)^{n}}{n(\ln n)^{2}}\). Although the terms as a whole converge by the alternating series test, the summation using absolute values of the terms, \(\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{2}}\), diverges.This unique point of \(x = -1\) allows for convergence due to the cancelling effects of alternating positive and negative terms, demonstrating why assessing the nature of convergence is essential in understanding series behavior.