Problem 34 - Essential Calculus Early Transcendentals

Question

Suppose that the power series \(\Sigma c_{n}(x-a)^{n}\) satisfies \(c_{n} \neq 0\) for all \(n .\) Show that if \(\lim _{n \rightarrow \infty}\left|c_{n} / c_{n+1}\right|\) exists, then it is equal to the radius of convergence of the power series.

Step-by-Step Solution

Answer

The limit \( \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| \) equals the radius of convergence.

Step 1: Understanding the Problem

We are given a power series of the form \( \sum c_n (x-a)^n \) and are tasked with proving that the radius of convergence can be found if the limit \( \lim _{n \rightarrow \infty}\left|\frac{c_n}{c_{n+1}}\right| \) exists.

Step 2: Using the Formula for Radius of Convergence

The radius of convergence \( R \) of a power series \( \sum c_n (x-a)^n \) can be found using the formula: \( R = \limsup_{n \to \infty} \left| \frac{1}{a_n} \right|^{1/n} \), where \( a_n = c_n(x-a)^n \). However, in practice, for ratio tests, it simplifies directly to \( R = \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| \) if this limit exists.

Step 3: Calculate the Limit

Since we assume that \( \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| \) exists, we denote this limit as \( L \). By the theory of power series convergence, this \( L \) directly gives us the radius of convergence \( R \).

Step 4: Conclude the Radius of Convergence

After calculating \( L \), the radius of convergence \( R \) is equal to this limit. Hence, we conclude that \( \lim _{n \rightarrow \infty}\left| \frac{c_n}{c_{n+1}}\right| \) is indeed equal to the radius of convergence of the power series.

Key Concepts

Radius of Convergence

The radius of convergence is a key concept when dealing with power series. It essentially tells us the range of values of the variable, often denoted as \(x\), for which the power series converges. Consider a power series of the form \( \sum c_n (x-a)^n \). The series converges within a certain distance from the center \(a\). This distance is what we call the radius of convergence, denoted \(R\).

• If \(|x-a| < R\), the series converges.
• If \(|x-a| > R\), the series diverges.
• If \(|x-a| = R\), convergence must be checked separately.

The radius of convergence can be found using various tests, one being the ratio test. Knowing \(R\) allows us to understand where our series will provide valid, meaningful answers.

Ratio Test

The ratio test is a powerful tool for finding the radius of convergence of a power series. It focuses on the limits of the ratios of successive terms. For a series \( \sum c_n (x-a)^n \), we often use the ratio of consecutive terms: \( \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} \). Simplifying this gives \( \frac{c_{n+1}}{c_n} (x-a) \).

To apply the ratio test to find the radius of convergence:

• Compute \( \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| \).
• Let this limit be \(L\). According to the test, \( R = \frac{1}{L} \).

If \( R \) is finite, the series converges for \(|x-a| < R\). If the limit \(L\) is zero, the series converges for all \(x\), indicating an infinite radius of convergence.

Limits in Calculus

Limits are fundamentally important in calculus and crucial for understanding convergence in power series. A limit describes the behavior of a function or sequence as it approaches a certain point. In the context of power series, we are interested in the limit of ratios of successive coefficients, \( \lim_{n \to \infty} \left| \frac{c_n}{c_{n+1}} \right| \).

This limit helps us determine the radius of convergence:\

• If the limit exists and is \(L\), the radius of convergence \(R\) is simply \(L\).
• If the limit is infinite or does not exist, convergence may not easily be determined.

Understanding limits allows us to make strong conclusions about the behavior of power series over different intervals. They serve as the basis for much of the analysis in calculus.