Question

A power series is given. (a) Find the radius of convergence. (b) Find the interval of convergence. $$\sum_{n=0}^{\infty} \frac{n}{3^{n}} x^{n}$$

Short answer

Radius of convergence is 3. Interval of convergence is \((-3, 3)\).

Step-by-step solution

Identify the General Term

The given power series is \( \sum_{n=0}^{\infty} \frac{n}{3^{n}} x^{n} \). The general term of this series is \( a_n = \frac{n}{3^{n}} x^{n} \).

Apply the Ratio Test to Find the Radius of Convergence

For the power series \( \sum a_n x^n \), use the ratio test formula: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute \( a_n = \frac{n}{3^n} x^n \) into the formula: \[ \lim_{n \to \infty} \left| \frac{\frac{n+1}{3^{n+1}} x^{n+1}}{\frac{n}{3^{n}} x^{n}} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot \frac{x}{3} \right| = \left| \frac{x}{3} \right| \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right) = \left| \frac{x}{3} \right|. \] For convergence, this limit must be less than 1: \( \left| \frac{x}{3} \right| < 1 \). This simplifies to \( |x| < 3 \). Therefore, the radius of convergence is \( R = 3 \).

Determine the Interval of Convergence

With \( R = 3 \), the interval is initially \(-3 < x < 3\). However, the interval of convergence must also consider the endpoints \( x = -3 \) and \( x = 3 \).

Test Endpoint \( x = -3 \)

Substitute \( x = -3 \) into the series: \[ \sum_{n=0}^{\infty} \frac{n}{3^{n}} (-3)^{n} = \sum_{n=0}^{\infty} n (-1)^{n}. \] This is an alternating series, but does not satisfy the alternating series test because \( \lim_{n \to \infty} n eq 0 \), hence it diverges.

Test Endpoint \( x = 3 \)

Substitute \( x = 3 \) into the series:\[ \sum_{n=0}^{\infty} \frac{n}{3^{n}} (3)^{n} = \sum_{n=0}^{\infty} n. \] This series clearly diverges because it sums all positive integers.

State the Final Interval of Convergence

Since the series diverges at both endpoints \( x = -3 \) and \( x = 3 \), the interval of convergence remains \((-3, 3)\), not including the endpoints.

Key concepts

Radius of Convergence

In the realm of power series, the radius of convergence is an essential concept. It determines the value of \( x \) for which a series converges. Given a power series \( \sum a_n x^n \), we seek a real number, \( R \), which defines this radius. The goal is to find values of \( x \) such that the series converges, specifically when \( |x| < R \). Outside this range, the series diverges. There are cases where a power series converges for specific scattered values of \( x \), but most commonly, finding this radius makes solving practical problems easier. To determine \( R \), we apply the Ratio Test, which is particularly well-suited for this purpose. By taking the limit\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|,\]we can derive \[ R = \frac{1}{\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|}.\] If the Ratio Test successfully delivers \( R = 3 \) for our example series, it implies convergence occurs whenever \( |x| < 3 \). This creates a circle of convergence with radius 3, centered at the origin on the real number line.

Interval of Convergence

The interval of convergence goes hand in hand with the radius of convergence. While the radius tells us about convergence at the center, the interval defines where along the entire real number line a power series converges. It is deduced by assessing the endpoints derived from the inequality \( |x| < R \). When you apply \( R = 3 \) to the initial interval \( -3 < x < 3 \), you get the possible interval but must consider what happens at \( x = -3 \) and \( x = 3 \).

Testing the endpoints is crucial as it verifies whether they should be included in the interval. For our series:

• The point \( x = -3 \) introduces the series \( \sum_{n=0}^{\infty} n(-1)^n \). Since this diverges, \( x = -3 \) must be excluded.
• At \( x = 3 \), the series \( \sum_{n=0}^{\infty} n \) also diverges, meaning \( x = 3 \) cannot be included either.

Consequently, the final interval of convergence strictly remains \((-3, 3)\), with open brackets, indicating that neither endpoint leads to convergence.

Ratio Test

The Ratio Test plays a fundamental role in determining convergence for power series, especially when calculating the radius of convergence. It involves comparing the terms of the series as they grow larger.For a series \( \sum a_n x^n \), the Ratio Test inspects the limit:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|. \]Here's a simple step-by-step on using this test:

• Determine \( a_n \): Recognize the general term of the series.
• Compute the Ratio: Form \( \frac{a_{n+1}}{a_n} \) utilizing the expressions for \( a_n \) and \( a_{n+1} \).
• Simplify: Often resulting in a simpler ratio of \( x \).
• Limit Calculation: Carry out the limit as \( n \rightarrow \infty \).

If \( \left| \frac{x}{3} \right| < 1 \), it simplifies to \( |x| < 3 \), resulting in the radius \( R = 3 \). It's a reliable method for analyzing series convergence, as it gives a numerical boundary for the valid \( x \) values.