Problem 31 - 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? $$ \sum_{n=1}^{\infty} \frac{(4 x-5)^{2 n+1}}{n^{3 / 2}} $$

Step-by-Step Solution

Answer

(a) Radius of convergence is 0.25, interval is (1, 1.5]. (b) Converges absolutely for 1 < x ≤ 1.5. (c) Converges conditionally at x = 1.

Step 1: Identify the series form

The given series is \( \sum_{n=1}^{\infty} \frac{(4x-5)^{2n+1}}{n^{3/2}} \). This is a power series in the form \( \sum_{n=1}^{\infty} c_n (4x-5)^{2n+1} \) with \( c_n = \frac{1}{n^{3/2}} \).

Step 2: Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test. Consider the general term \( a_n = \frac{(4x-5)^{2n+1}}{n^{3/2}} \). Compute the limit: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(4x-5)^{2n+3}}{(n+1)^{3/2}} \cdot \frac{n^{3/2}}{(4x-5)^{2n+1}} \right|. \]Simplify:\[ L = \lim_{n \to \infty} (4x-5)^2 \cdot \left( \frac{n}{n+1} \right)^{3/2}. \]

Step 3: Simplify the limit calculation

Notice that \( \left( \frac{n}{n+1} \right)^{3/2} \to 1 \) as \( n \to \infty \). Thus, the limit \( L = (4x-5)^2 \). For convergence, we require \( L < 1 \), so:\[ (4x-5)^2 < 1. \]

Step 4: Solve the inequality

Solve \( (4x-5)^2 < 1 \): \[ -1 < 4x-5 < 1. \]Add 5 to each part: \[ 4 < 4x < 6. \]Divide all sides by 4: \[ 1 < x < 1.5. \]Thus, the radius of convergence \( R \) is \( 0.25 \) and the interval is \((1, 1.5)\).

Step 5: Check convergence at endpoints

Examine \( x = 1 \) and \( x = 1.5 \).For \( x = 1 \): The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^{2n+1}}{n^{3/2}} \), which is conditionally convergent by the Alternating Series Test.For \( x = 1.5 \): The series becomes \( \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} \), which is convergent and absolutely convergent by the p-series test with \( p = \frac{3}{2} > 1 \).

Step 6: Conclude convergence types

(b) The series converges absolutely for \( 1 < x \leq 1.5 \) because at \( x = 1.5 \) the series is absolutely convergent.(c) The series converges conditionally at \( x = 1 \) because it converges but not absolutely.

Key Concepts

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) represents the coefficients of the series, \( x \) is a variable, and \( c \) is a constant known as the center of the series. In the series given in the exercise, \( \sum_{n=1}^{\infty} \frac{(4x-5)^{2n+1}}{n^{3/2}} \), the center is affected by the transformation \( 4x-5 \), which shifts the series, and it involves odd powers of the term due to \((2n+1)\). Power series are incredibly useful because they allow functions to be expressed as infinite polynomials, making them easier to manipulate and analyze. They serve as the foundation for many higher-level mathematical concepts and applications. Understanding power series enables you to work with functions in a manageable and approximable form over certain intervals around the center \(c\).
To effectively utilize power series, it's critical to determine their convergence properties. This is where tests like the ratio test become handy, as they help ascertain the series' radius and interval of convergence, which determine where the series converges to a function.

Ratio Test

The ratio test is an essential tool in calculus for determining the convergence of infinite series. When we're given a series \( \sum a_n \), the ratio test involves calculating the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). The result of this limit tells us about the series' convergence:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L = \infty \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

For the series \( \sum_{n=1}^{\infty} \frac{(4x-5)^{2n+1}}{n^{3/2}} \), applying the ratio test leads to finding \( L = (4x-5)^2 \). By setting \( (4x-5)^2 < 1 \), we can determine the values of \( x \) for which the series converges. This process helps us find both the radius of convergence and the interval in which the series behaves like a convergent power series. Mastering the ratio test is crucial for tackling problems involving series and their convergence.

Interval of Convergence

The interval of convergence is a range of \( x \)-values where a power series converges to a finite value. It’s derived after determining the radius of convergence, which is the distance from the center \( c \) to the boundaries of the interval on the number line. For our series, solving the inequality \( (4x-5)^2 < 1 \), we find the interval \((1, 1.5)\). Inside this interval, the power series converges.
Sometimes, to find the complete interval of convergence, it's necessary to check convergence at the endpoints separately. This might involve applying additional tests, such as the alternating series test or the p-series test, to check absolute or conditional convergence at these points. Properly understanding the interval of convergence allows you to know exactly where a power series will provide valid and accurate results.

Absolute Convergence

A series is said to be absolutely convergent if the series of the absolute values of its terms also converges. This means if \( \sum a_n \) converges, then checking for \( \sum |a_n| \) should also give a convergent series. For the series in the exercise, we've noted that for \( x = 1.5 \), the series \( \sum \frac{1}{n^{3/2}} \) is absolutely convergent due to the p-series test with \( p = \frac{3}{2} > 1 \).
Absolute convergence is a stronger form of convergence compared to conditional convergence. If a series converges absolutely, it implies convergence in the usual sense, no matter how you might rearrange the terms—invaluable for many mathematical analyses. Recognizing absolute convergence helps when dealing with more complex mathematical operations such as integration and differentiation of series.

Conditional Convergence

Conditional convergence occurs when a series converges, but not absolutely. This nuanced concept is showcased in our series when examined for \( x = 1 \). Here, the series \( \sum \frac{(-1)^{2n+1}}{n^{3/2}} \) converges by the alternating series test, but not absolutely since \( \sum \frac{1}{n^{3/2}} \) does not converge.
Conditional convergence reveals that while the series achieves finite sums due to its arrangement, rearranging its terms could lead to a different, possibly divergent, sum. Therefore, subtle insights into the behavior of the series can be gleaned, often requiring more caution when performing manipulations like rearrangements. Understanding conditional convergence helps in recognizing which mathematical properties can be safely applied to a series.