Find The Interval Of Convergence Of The Series

Finding the Interval of Convergence of a Power Series

Determining the interval of convergence for a power series is a crucial step in understanding its behavior and applications. This process involves identifying the range of x-values for which the series converges, be it absolutely, conditionally, or diverges. This article will guide you through the process, exploring various methods and providing practical examples. We'll delve into the intricacies of the Ratio Test, the Root Test, and the consideration of endpoints to ensure a comprehensive understanding.

Understanding Power Series

Before diving into the methods, let's refresh our understanding of power series. A power series is an infinite series of the form:

∑_n=0^∞ c_n(x - a)ⁿ = c₀ + c₁(x - a) + c₂(x - a)² + ...

where:

• c_n are the coefficients of the series (constants).
• x is a variable.
• a is the center of the series (a constant).

The interval of convergence is the set of all x-values for which the power series converges. This interval is often centered around 'a' and has a radius of convergence, 'R'.

Methods for Determining the Interval of Convergence

The most common methods for determining the interval of convergence are the Ratio Test and the Root Test. Let's examine each:

1. The Ratio Test

The Ratio Test is frequently used to determine the radius of convergence. It states that for a series ∑a_n:

• If lim_n→∞ |a_n+1/a_n| = L < 1, the series converges absolutely.
• If lim_n→∞ |a_n+1/a_n| = L > 1, the series diverges.
• If lim_n→∞ |a_n+1/a_n| = L = 1, the test is inconclusive.

Applying the Ratio Test to Power Series:

When applying the Ratio Test to a power series, we examine the limit:

lim_n→∞ |c_n+1(x - a)ⁿ⁺¹ / c_n(x - a)ⁿ| = |x - a| lim_n→∞ |c_n+1 / c_n|

Let's denote lim_n→∞ |c_n+1 / c_n| = 1/R. Then the series converges absolutely when:

|x - a| (1/R) < 1 => |x - a| < R

This means the series converges absolutely within the interval (a - R, a + R). The value 'R' is the radius of convergence.

2. The Root Test

The Root Test provides an alternative approach, particularly useful when dealing with series containing exponential terms or factorials. It states that for a series ∑a_n:

• If lim_n→∞ |a_n|^1/n = L < 1, the series converges absolutely.
• If lim_n→∞ |a_n|^1/n = L > 1, the series diverges.
• If lim_n→∞ |a_n|^1/n = L = 1, the test is inconclusive.

Applying the Root Test to Power Series:

Applying the Root Test to a power series, we analyze:

lim_n→∞ |c_n(x - a)ⁿ|^1/n = |x - a| lim_n→∞ |c_n|^1/n

Let's assume lim_n→∞ |c_n|^1/n = 1/R. The series converges absolutely when:

|x - a| (1/R) < 1 => |x - a| < R

Again, R represents the radius of convergence, and the series converges absolutely within (a - R, a + R).

Checking the Endpoints

The Ratio and Root Tests determine the radius of convergence, giving us an open interval. However, the series might also converge at one or both endpoints of this interval. To determine this, we must check the convergence at x = a - R and x = a + R separately using other convergence tests like the Alternating Series Test, p-series test, Integral Test, Comparison Test, Limit Comparison Test, etc.

Examples

Let's illustrate the process with a few examples:

Example 1: Find the interval of convergence for the series ∑_n=1^∞ (xⁿ)/n.

Solution:

We use the Ratio Test:

lim_n→∞ |(xⁿ⁺¹/(n+1)) / (xⁿ/n)| = lim_n→∞ |x| * (n/(n+1)) = |x|

The series converges absolutely when |x| < 1, meaning the radius of convergence is R = 1. The interval is (-1, 1).

Now we check the endpoints:

• x = -1: The series becomes ∑_n=1^∞ (-1)ⁿ/n, which converges by the Alternating Series Test.
• x = 1: The series becomes ∑_n=1^∞ 1/n, which is the harmonic series and diverges.

Therefore, the interval of convergence is [-1, 1).

Example 2: Find the interval of convergence for the series ∑_n=0^∞ (x - 2)ⁿ / (n!)

Solution:

Using the Ratio Test:

lim_n→∞ |((x - 2)ⁿ⁺¹/(n+1)!) / ((x - 2)ⁿ/n!)| = lim_n→∞ |x - 2| / (n + 1) = 0

Since the limit is 0 for all x, the series converges absolutely for all x. Therefore, the interval of convergence is (-∞, ∞). This is an example of a series with an infinite radius of convergence.

Example 3: Find the interval of convergence for the series ∑_n=0^∞ (n!(x+3)^n)/2^n

Solution:

Using the Ratio Test:

lim_n→∞ | [(n+1)!(x+3)^(n+1)/2^(n+1)] / [n!(x+3)^n/2^n] | = lim_n→∞ |(n+1)(x+3)/2|

For convergence, we need |(n+1)(x+3)/2| < 1. This limit tends to infinity unless x+3 = 0 which gives x=-3. In this case the limit is 0, which means the series converges only for x = -3. The interval of convergence is therefore {-3}.

These examples highlight the importance of systematically applying the Ratio or Root Test, followed by a thorough examination of the endpoints to completely define the interval of convergence. Remember that the choice between the Ratio and Root Test often depends on the structure of the series, with the Ratio Test being generally preferred for its relative simplicity in many cases. However, the Root Test can be more effective when dealing with series involving n-th roots or complex terms. Understanding these tests and their applications is fundamental to mastering power series analysis. Always remember to check the endpoints; they can significantly impact the final interval of convergence.