Problem 9 - 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? $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n \sqrt{n} 3^{n}} $$

Step-by-Step Solution

Answer

(a) Radius: 3; interval: \([-3, 3)\). (b) Absolutely: \((-3, 3)\). (c) Conditionally: \(x = -3\).

Step 1: Identify the series components

The given series is \( \sum_{n=1}^{\infty} \frac{x^n}{n \sqrt{n} 3^n} \). Here, the general term is \( a_n = \frac{x^n}{n \sqrt{n} 3^n} \).

Step 2: Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{(n+1)\sqrt{n+1}3^{n+1}} \cdot \frac{n\sqrt{n}3^n}{x^n} \right|\]This simplifies to:\[\lim_{n \to \infty} \left| \frac{x}{3} \cdot \frac{n \sqrt{n}}{(n+1)\sqrt{n+1}} \right| = \left| \frac{x}{3} \right| \cdot \lim_{n \to \infty} \frac{n \sqrt{n}}{(n+1)\sqrt{n+1}}\]As \(n \to \infty\), \(\frac{n \sqrt{n}}{(n+1)\sqrt{n+1}} \to 1\). Thus:\[\lim = \left| \frac{x}{3} \right|\] To converge, it must satisfy \( \left| \frac{x}{3} \right| < 1 \).

Step 3: Solve for the radius of convergence

From \( \left| \frac{x}{3} \right| < 1 \), we have \( |x| < 3 \). Therefore, the radius of convergence is 3. The interval is initially \((-3, 3)\).

Step 4: Check endpoints for interval convergence

Now, consider the endpoints \(x = -3\) and \(x = 3\):- For \(x = -3\), the series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n\sqrt{n}}\). This is the alternating series, which converges.- For \(x = 3\), the series becomes \(\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}\), which diverges by the Limit Comparison Test (by comparing with \(\sum \frac{1}{n^{3/2}}\)).Thus, the interval of convergence is \([-3, 3)\).

Step 5: Determine absolute convergence

The series converges absolutely within \(|x| < 3\). For the endpoint \(x = -3\), check absolute convergence:\[\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}}\] also diverges.Therefore, absolute convergence is strictly within \((-3, 3)\).

Step 6: Determine conditional convergence

For \(x = -3\), as previously found, \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n\sqrt{n}} \) converges due to the Alternating Series Test but does not converge absolutely. Therefore, it converges conditionally only at \(x = -3\).

Key Concepts

Ratio Test

The Ratio Test is a valuable tool in determining the radius of convergence of a series. Let's examine it using our series, \[ \sum_{n=1}^{\infty} \frac{x^n}{n \sqrt{n} 3^n}. \]The Ratio Test involves taking the limit of the absolute value of the ratio of consecutive terms, which can be expressed mathematically as:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]For our series, substituting the general terms, we simplify and find the crucial part:\[\lim_{n \to \infty} \left| \frac{x}{3} \right| \cdot \lim_{n \to \infty} \frac{n \sqrt{n}}{(n+1)\sqrt{n+1}}. \]As \(n\) becomes very large, \(\frac{n \sqrt{n}}{(n+1)\sqrt{n+1}}\) approaches 1. Therefore, the limit simplifies to \(\left| \frac{x}{3} \right| \).To ensure convergence, this must be less than 1, resulting in \( |x| < 3 \). This condition tells us the series' radius of convergence is 3.

Interval of Convergence

The interval of convergence identifies the set of \(x\) values over which a series converges. After finding the radius of convergence as 3, the interval initially becomes \((-3, 3)\). However, endpoints may affect convergence, necessitating additional testing. For \(x = -3\), substituting into the series yields:\[\sum_{n=1}^{\infty} \frac{(-1)^n}{n \sqrt{n}}, \]an alternating series converging by the Alternating Series Test.For \(x = 3\), substituting gives:\[\sum_{n=1}^{\infty} \frac{1}{n \sqrt{n}}, \]which diverges by comparison with a divergent \(p\)-series \(\sum \frac{1}{n^{3/2}} \).Thus, the true interval of convergence turns out to be \([-3, 3)\), as the series converges at \(x = -3\) but not at \(x = 3\).

Absolute Convergence

Absolute convergence occurs when the series formed by taking the absolute value of each term also converges. For our series:\[ \sum_{n=1}^{\infty} \frac{x^n}{n \sqrt{n} 3^n}, \]absolute convergence happens within \(|x| < 3\). This stems from the Ratio Test indicating convergence within the same range.At the endpoint \(x = -3\), we looked at the series:\[\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n}},\]which diverges. Consequently, despite conditional convergence being present here, absolute convergence is limited to less than 3 on both sides, thus within \((-3, 3)\).

Conditional Convergence

Conditional convergence arises when a series converges, but not absolutely. This is a subtler form of convergence that typically appears in alternating series. For \(x = -3\), the series:\[\sum_{n=1}^{\infty} \frac{(-1)^n}{n\sqrt{n}}\]converges by the Alternating Series Test.However, the test for absolute convergence, as discussed earlier, does not hold for this endpoint. The series formed by the absolute values diverges, meaning it converges conditionally.Ultimately, in our specific scenario, conditional convergence only occurs at \(x = -3\), identified through the analysis of convergence tests and behaviors of the series.