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Find the radius of convergence of each power series. $$ 1-\frac{x}{3}+\frac{x^{2}}{3^{2}}-\frac{x^{3}}{3^{3}}+\cdots+\frac{(-1)^{n} x^{n}}{3^{n}}+\cdots $$

Short Answer

Expert verified
The radius of convergence is 3.

Step by step solution

01

Identify the general term of the series

The series is given by \(1 - \frac{x}{3} + \frac{x^2}{3^2} - \frac{x^3}{3^3} + \cdots + \frac{(-1)^n x^n}{3^n} + \cdots\). The general term of the series is \(a_n = \frac{(-1)^n x^n}{3^n}\).
02

Apply the Ratio Test for convergence

The Ratio Test states that for the series \(\sum a_n\), the radius of convergence \(R\) can be found by evaluating \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
03

Compute \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)

Calculate the ratio \(\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} x^{n+1} / 3^{n+1}}{(-1)^n x^n / 3^n} = \frac{(-1)^{n+1} x^{n+1}}{3 x^n / 3^n (-1)^n}\). Simplifying gives \(\frac{(-1)x}{3}\), leading to \(|x|/3\). Thus, \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|}{3}\).
04

Determine the radius of convergence \(R\)

For convergence using the Ratio Test, we need \(\frac{|x|}{3} < 1\), which implies \(|x| < 3\). Therefore, the radius of convergence is \(R = 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test for Convergence
The Ratio Test is a powerful tool for determining the convergence of a series. This test focuses on the limit of the ratio of successive terms. Given a series \(\sum a_n\), you test convergence by evaluating \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).

If this limit is less than 1, the series is absolutely convergent; if greater than 1, the series diverges. It’s especially handy for series whose terms involve factorials or exponential functions. In our example, the series \(1 - \frac{x}{3} + \frac{x^2}{3^2} - \cdots\) uses the Ratio Test. We find the general term as \(a_n = \frac{(-1)^n x^n}{3^n}\).

Using the Ratio Test, we calculate \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|}{3}\). For convergence, \(\frac{|x|}{3} < 1\) implies \(|x| < 3\), resulting in a radius of convergence \(R = 3\).
Understanding Power Series
A power series is a series of the form \(\sum_{n=0}^{\infty} c_n x^n\), where \(c_n\) are coefficients and \(x\) is a variable. This series can be understood like a polynomial of infinite degree. The key detail is how \(x\) behaves within it to influence convergence.

Each power series is about a specific center, often at \(x=0\). This is called the center of the series. The set of all \(x\) within which this series converges is determined by the radius of convergence, \(R\).

In our example, the series involves terms of the form \(\frac{(-1)^n x^n}{3^n}\), showcasing that each term's coefficient \(c_n = \frac{(-1)^n}{3^n}\). To determine where the series converges, you utilize tools like the Ratio Test, which examine how \(x\) can vary without making the series diverge.
Convergence of Power Series
Convergence refers to whether the sum of a series approaches a finite limit as more terms are added. For a power series, convergence is primarily about when the sum reaches this limit as the variable \(x\) changes.

The Radius of Convergence \(R\) is crucial here, delineating the interval \(-R < x < R\) within which the series converges. Outside this interval, the series is likely to diverge.
  • Inside the radius \(R\), the series sums to a specific value.
  • Exactly at \(x = R\) and \(x = -R\), things can be trickier, and each case requires a unique check for convergence.

In our example, the power series \(1 - \frac{x}{3} + \frac{x^2}{3^2} - \cdots\) converges when \(|x| < 3\). As calculated, its radius \(R = 3\) shows that within this range, you can be confident of convergence. Thus, convergence is about figuring the limits within which the power series behaves nicely.

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