Determining Radius And Interval Of Convergence For Power Series

In the realm of mathematical analysis, power series play a pivotal role, offering a way to represent functions as infinite sums of terms involving powers of a variable. Understanding the convergence behavior of these series is crucial for various applications, including approximating functions, solving differential equations, and exploring complex analysis. This article delves into the process of determining the radius and interval of convergence for power series, focusing on three specific examples. We will explore the application of the ratio test and endpoint analysis to rigorously define the region where these series converge.

Understanding Power Series Convergence

Before diving into the examples, let's establish a firm understanding of power series convergence. A power series is an infinite series of the form:

$$\sum_{n=0}^{\infty} c_n (x-a)^n$$

where $c_n$ are the coefficients, $x$ is the variable, and $a$ is the center of the series. The convergence of a power series hinges on the values of $x$. A fundamental theorem states that for any power series, there exists a non-negative number $R$ (the radius of convergence) such that the series converges if $|x - a| < R$ and diverges if $|x - a| > R$. The interval of convergence is the interval of all $x$ values for which the series converges. It is centered at $a$ and has a length of $2R$, but it's crucial to examine the endpoints $a - R$ and $a + R$ separately, as the series may converge at one or both endpoints, or diverge at both.

To determine the radius of convergence, the ratio test is a powerful tool. It involves calculating the limit:

$$L = \lim_{n \to \infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n} \right| = \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| |x-a|$$

If $L < 1$, the series converges; if $L > 1$, the series diverges; and if $L = 1$, the test is inconclusive. From this, we can often extract the radius of convergence $R$. Once we have $R$, we must test the endpoints $x = a - R$ and $x = a + R$ individually using other convergence tests, such as the comparison test, the alternating series test, or the integral test, to determine whether they are included in the interval of convergence.

a. ${\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}}}$

Let's first discuss the power series ${\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}}}$. Finding the radius of convergence is the initial step in understanding the behavior of this series. We will employ the ratio test, a cornerstone technique for analyzing series convergence. The ratio test involves examining the limit of the ratio of consecutive terms as n approaches infinity. This limit, if it exists, provides crucial information about the series' convergence properties. Specifically, we consider the limit:

$$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$

where $a_n = \frac{x^n}{\sqrt{n}}$. Substituting this into the limit, we get:

$$L = \lim_{n \to \infty} \left| \frac{x^{n+1}}{\sqrt{n+1}} \cdot \frac{\sqrt{n}}{x^n} \right|$$

Simplifying the expression inside the limit, we have:

$$L = \lim_{n \to \infty} |x| \cdot \sqrt{\frac{n}{n+1}}$$

The term $\sqrt{\frac{n}{n+1}}$ approaches 1 as n approaches infinity. Therefore, the limit simplifies to:

$$L = |x|$$

For the series to converge, the ratio test dictates that L must be less than 1. Thus, we have the condition:

$$|x| < 1$$

This inequality directly gives us the radius of convergence, R. In this case, R = 1. This means the series converges for all x values within the interval (-1, 1). However, the ratio test provides no information about the convergence at the endpoints, x = -1 and x = 1. We must investigate these endpoints separately.

Now, let's examine the endpoints to determine the interval of convergence. When x = 1, the series becomes:

$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$$

This series is a p-series with p = 1/2. Since p ≤ 1, the p-series diverges. Therefore, the series diverges at x = 1.

Next, consider the case when x = -1. The series becomes:

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$$

This is an alternating series. To determine its convergence, we can apply the Alternating Series Test. The Alternating Series Test requires two conditions to be met: the absolute values of the terms must decrease monotonically, and the limit of the terms must approach zero. In this case, the terms are $\frac{1}{\sqrt{n}}$. These terms clearly decrease as n increases, and the limit as n approaches infinity is zero. Therefore, the alternating series converges at x = -1.

Combining these results, we find that the interval of convergence for the series ${\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n}}}$ is [-1, 1). The series converges for all x values greater than or equal to -1 and strictly less than 1. The inclusion of -1 and the exclusion of 1 in the interval represent the nuanced convergence behavior at the endpoints, which is a critical aspect of power series analysis.

c. ${\sum_{n=0}^{\infty} n x^{2n}}$

Now, let’s consider the power series ${\sum_{n=0}^{\infty} n x^{2n}}$. This series features a slightly different structure than the previous example, with terms involving x raised to the power of 2n. Despite this difference, the fundamental approach to finding the radius of convergence remains the same: the ratio test. We begin by defining the general term of the series as:

$$a_n = n x^{2n}$$

Applying the ratio test, we consider the limit:

$$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)x^{2(n+1)}}{n x^{2n}} \right|$$

Simplifying the expression inside the limit, we obtain:

$$L = \lim_{n \to \infty} \left| \frac{(n+1)x^{2n+2}}{n x^{2n}} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n} x^2 \right|$$

As n approaches infinity, the fraction $\frac{n+1}{n}$ approaches 1. Thus, the limit simplifies to:

$$L = |x^2| = x^2$$

For the series to converge, the ratio test requires that L be less than 1. Therefore, we have the condition:

$$x^2 < 1$$

Taking the square root of both sides, we get:

$$|x| < 1$$

This inequality reveals that the radius of convergence, R, is 1. The series converges for all x values within the interval (-1, 1). As before, we must now investigate the endpoints, x = -1 and x = 1, to determine the interval of convergence.

Let's examine the endpoint x = 1. Substituting x = 1 into the series, we get:

$$\sum_{n=0}^{\infty} n(1)^{2n} = \sum_{n=0}^{\infty} n$$

This series is a simple sum of positive integers. It clearly diverges because the terms do not approach zero as n approaches infinity. The divergence at x = 1 is quite straightforward.

Next, consider the endpoint x = -1. Substituting x = -1 into the series, we obtain:

$$\sum_{n=0}^{\infty} n(-1)^{2n} = \sum_{n=0}^{\infty} n$$

Since (-1)^(2n) is always 1, this series is identical to the series obtained when x = 1. Therefore, it also diverges. The divergence at x = -1 mirrors the divergence at x = 1, which is an important observation.

Combining these results, we conclude that the interval of convergence for the series ${\sum_{n=0}^{\infty} n x^{2n}}$ is (-1, 1). The series converges for all x values strictly between -1 and 1. The exclusion of both endpoints from the interval highlights a crucial aspect of power series convergence: the endpoints must be examined individually, and their behavior can significantly impact the overall interval of convergence.

e. ${\sum_{n=0}^{\infty} n^2 (x-4)^n}$

Finally, let’s analyze the power series ${\sum_{n=0}^{\infty} n^2 (x-4)^n}$. This series is centered at x = 4, which introduces a shift in the convergence interval. The process of finding the radius and interval of convergence remains consistent: we begin with the ratio test. Defining the general term of the series as:

$$a_n = n^2 (x-4)^n$$

we apply the ratio test by considering the limit:

$$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(n+1)^2 (x-4)^{n+1}}{n^2 (x-4)^n} \right|$$

Simplifying the expression inside the limit, we get:

$$L = \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} (x-4) \right|$$

As n approaches infinity, the fraction $\frac{(n+1)^2}{n^2}$ approaches 1. To see this, we can expand the numerator and divide both numerator and denominator by $n^2$:

$$\lim_{n \to \infty} \frac{(n+1)^2}{n^2} = \lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^2} = \lim_{n \to \infty} \left( 1 + \frac{2}{n} + \frac{1}{n^2} \right) = 1$$

Thus, the limit simplifies to:

$$L = |x-4|$$

For the series to converge, the ratio test requires that L be less than 1. Therefore, we have the condition:

$$|x-4| < 1$$

This inequality indicates that the radius of convergence, R, is 1. The series converges for all x values within 1 unit of the center x = 4. This gives us the interval (3, 5). Now, we must examine the endpoints, x = 3 and x = 5, to determine the complete interval of convergence.

Let's consider the endpoint x = 5. Substituting x = 5 into the series, we get:

$$\sum_{n=0}^{\infty} n^2 (5-4)^n = \sum_{n=0}^{\infty} n^2$$

This series is a sum of squares of integers. It clearly diverges because the terms do not approach zero as n approaches infinity. The divergence at x = 5 is evident.

Next, consider the endpoint x = 3. Substituting x = 3 into the series, we obtain:

$$\sum_{n=0}^{\infty} n^2 (3-4)^n = \sum_{n=0}^{\infty} n^2 (-1)^n$$

This is an alternating series, but the terms $n^2$ do not approach zero as n approaches infinity. Therefore, the series diverges by the Divergence Test. The Divergence Test states that if the terms of a series do not approach zero, then the series must diverge. In this case, the terms $n^2 (-1)^n$ oscillate between large positive and negative values, never approaching zero, which confirms the divergence.

Combining these results, we find that the interval of convergence for the series ${\sum_{n=0}^{\infty} n^2 (x-4)^n}$ is (3, 5). The series converges for all x values strictly between 3 and 5. The exclusion of both endpoints from the interval emphasizes the importance of endpoint analysis in determining the complete convergence behavior of a power series.

Conclusion

In this exploration, we have demonstrated the process of finding the radius and interval of convergence for three distinct power series. The ratio test serves as a foundational tool for determining the radius of convergence, while endpoint analysis, often involving tests like the p-series test, alternating series test, or divergence test, is crucial for defining the complete interval of convergence. Understanding these concepts is essential for working with power series and their applications in various areas of mathematics and physics. Through careful application of these techniques, we can rigorously define the regions where power series converge, enabling us to leverage their power in approximating functions and solving complex problems.