Problem 13 - Essential Calculus Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{(-3)^{n}}{n \sqrt{n}} x^{n}$$

Step-by-Step Solution

Answer

Radius of convergence is \( \frac{1}{3} \); interval is \( (-\frac{1}{3}, \frac{1}{3}) \).

Step 1: Identify Given Series

The series given is \( \sum_{n=1}^{\infty} \frac{(-3)^{n}}{n \sqrt{n}} x^{n} \). Our task is to find the radius of convergence \( R \) and the interval of convergence.

Step 2: Use the Ratio Test

To find the radius of convergence, use the Ratio Test. Consider the general term \( a_n = \frac{(-3)^{n}}{n \sqrt{n}} x^{n} \). Compute \( \left| \frac{a_{n+1}}{a_n} \right| \).

Step 3: Compute Ratio of Consecutive Terms

Calculate \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-3)^{n+1}}{(n+1) \sqrt{n+1}} x^{n+1} \cdot \frac{n \sqrt{n}}{(-3)^{n} x^{n}} \right| = \left| \frac{-3x \cdot n \sqrt{n}}{(n+1) \sqrt{n+1}} \right| \).

Step 4: Simplify the Ratio

Simplifying, \( \left| \frac{a_{n+1}}{a_n} \right| = |3x| \cdot \frac{n}{n+1} \cdot \sqrt{\frac{n}{n+1}} \). Use the approximation for large \( n \), which simplifies this to \( |3x| \).

Step 5: Apply the Ratio Test

Set the limit \( \lim_{n \to \infty} |3x| = |3x| \). The series converges if \( |3x| < 1 \), which implies \( |x| < \frac{1}{3} \). Thus, the radius of convergence \( R \) is \( \frac{1}{3} \).

Step 6: Determine Interval of Convergence

The interval of convergence is initially \( (-\frac{1}{3}, \frac{1}{3}) \). Check endpoints \( x = \pm\frac{1}{3} \) separately by substituting into the series and testing for convergence. \( x = \frac{1}{3} \) leads to divergent behavior, and \( x = -\frac{1}{3} \) also leads to divergence due to alternating harmonic nature.

Step 7: Conclusion on Convergence Interval

Since the series diverges at both endpoints \( x = \frac{1}{3} \) and \( x = -\frac{1}{3} \), the interval of convergence is \( (-\frac{1}{3}, \frac{1}{3}) \).

Key Concepts

Interval of Convergence

The interval of convergence of a power series is crucial as it tells us where the series converges to a finite value. To find this interval, we first need to find the radius of convergence, which provides us with a range around the center point where the series converges.

For the series given, after determining the radius of convergence using the Ratio Test, we start with an open interval centered at zero:

• The series converges for any value of x such that \( |x| < \text{radius of convergence} \).
• Here, the radius is \( \frac{1}{3} \), so the initial interval is \((-\frac{1}{3}, \frac{1}{3})\).

Next, it's important to check the endpoints of this interval to see if they should be included, this can change the open interval to a closed or half-open interval. If substitution of the endpoints into the series leads to convergence, they are included. Otherwise, they're excluded. In this case, neither endpoint converges, so the interval remains open.

Ratio Test

The Ratio Test is a valuable tool to find the radius of convergence for power series. It involves analyzing the limit of the absolute value of consecutive terms in a series as the series progresses to infinity.

For a general term \(a_n\) in the series, the Ratio Test considers:

• The next term, \(a_{n+1}\), divided by \(a_n\).
• The limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

If this limit is less than 1, the series converges for those values of \(x\). Specifically,

• The series converges when this limit yields \( |3x| < 1 \).
• Solving \( |3x| < 1 \) gives an interval of \( |x| < \frac{1}{3} \).

Ultimately, the radius of convergence is the result from this test, and it dictates the range for convergence in relation to the absolute value of \(x\).

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where each term is a function of \(x\). This series is centered at \(c\), which in the exercise provided is zero. A crucial aspect of power series is determining where they converge.

The span around the center where the series converges is determined by:

• A radius of convergence \(R\), within which the series can sum to a finite number.
• An interval of convergence derived from \(R\), evaluated further to include or exclude endpoints depending on convergence tests.

In the exercise, the power series is presented with the general form and the need to compute or apply other tests such as the Ratio Test to find convergent intervals, which ultimately enhances our understanding of the series' behavior.

Alternating Series

An alternating series is when the terms in a series alternate between positive and negative. This gives the series a zig-zag behavior as it progresses. Alternating series can converge differently than non-alternating series, often with unique properties such as error bounds in their approximation, similar to a harmonic series but with alternating signs.

In this scenario:

• The series \( \sum_{n=1}^{\infty} \frac{(-3)^{n}}{n \sqrt{n}} x^{n} \) presents alternating signs due to the term \((-3)^n\).
• Such behavior necessitates careful analysis at the endpoints, often leading to divergence if individual terms don't tend to zero or if the series doesn't meet requisite alternating series conditions.

In our case, even though the endpoints were checked, the nature of the alternating terms contributed to the divergence at both \(x = \frac{1}{3}\) and \(x = -\frac{1}{3}\), hence maintaining an open interval of convergence.