Problem 11 - Calculus: Early Transcendentals

Question

Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 1}^{\infty} \frac {( - 1)^n 4^n}{\sqrt{n}} x^n \)

Step-by-Step Solution

Answer

Radius of convergence is \( \frac{1}{4} \) and interval of convergence is \([-\frac{1}{4}, \frac{1}{4})\).

Step 1: Identify the Format of the Power Series

The given series is: \[\sum_{n = 1}^{\infty} \frac{(-1)^n 4^n}{\sqrt{n}} x^n\]This is a power series in the form \( \sum a_n x^n \) with \( a_n = \frac{(-1)^n 4^n}{\sqrt{n}} \). We are asked to find the radius and interval of convergence for this series.

Step 2: Apply the Root Test

To find the radius of convergence, apply the root test. Compute \[\lim_{n \to \infty} \sqrt[n]{|a_n|}\].Substitute the term:\[\sqrt[n]{\left| \frac{(-1)^n 4^n}{\sqrt{n}} \right|} = \sqrt[n]{4^n} \cdot \sqrt[n]{\frac{1}{\sqrt{n}}} = 4 \cdot (n^{-1/(2n)})\].As \( n \to \infty \), \( n^{-1/(2n)} \to 1 \), hence the limit is \[4\].

Step 3: Determine the Radius of Convergence

Since the limit from the root test is 4, the radius of convergence \( R \) is given by \[R = \frac{1}{4}\].

Step 4: Test for Convergence at Endpoints

The radius of convergence from Step 3 suggests that the interval of convergence centered around 0 is \[(-\frac{1}{4}, \frac{1}{4})\].Check the endpoints separately:- **At \( x = \frac{1}{4} \)**:The series becomes:\[\sum_{n = 1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\],which is a conditionally convergent alternating series.- **At \( x = -\frac{1}{4} \)**:The series becomes:\[\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}}\],which is the harmonic series and divergent.

Step 5: State the Interval of Convergence

The results from the endpoint tests tell us the interval of convergence is \[[-\frac{1}{4}, \frac{1}{4})\].This means the series converges for \( -\frac{1}{4} \leq x < \frac{1}{4} \).

Key Concepts

Power Series

A power series is like an infinite polynomial where each term includes a variable raised to a power and multiplied by a coefficient. It looks like this:

• The general form is: \[ \sum_{n=0}^{\infty} a_n x^n, \]where \( a_n \) are coefficients.
• In this exercise, we have: \[ \sum_{n = 1}^{\infty} \frac{(-1)^n 4^n}{\sqrt{n}} x^n,\]which means the coefficients are \[ \frac{(-1)^n 4^n}{\sqrt{n}}.\]

Power series show up often in math because they can approximate functions very well. However, they might not be valid for all values of \( x \). That's why we need to find where the series "works," and that's called finding its radius and interval of convergence.
Understanding power series is key because they are foundational in calculus and can help in analyzing more complex functions.

Interval of Convergence

The interval of convergence helps us understand the range of values of \( x \) that makes the power series converge. This means it helps determine where the series (like an infinite sum) is defined accurately.

• First, find the radius of convergence. This tells us how far from the center, 0 in this case, we can go and still have a convergent series.
• Then, use this radius to find the interval. The series converges within this interval, often written as something like \( (a, b) \) or \( [a, b) \).
• Finally, it's important to check the endpoints separately because a power series can behave differently at the edges of the interval. For example, in this exercise, it converges at the left endpoint but not at the right.

Thus, when you talk about the interval of convergence, you ensure that the power series accurately approximates or defines a function within this specified range.

Root Test

The root test helps us determine the radius of convergence for a power series. By applying the root test, you can see mathematically where the series behaves well.

• For a power series \( \sum a_n x^n \), compute \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \).
• If the limit is \( L \), then the radius of convergence, \( R \), is given by \( R = \frac{1}{L} \). If \( L \) is zero, the radius is infinite, meaning convergence everywhere.

In our exercise, applying the root test gave us a limit, \( 4 \), hence the radius of convergence is \( \frac{1}{4} \).By using the root test, we have a reliable method to calculate where the complex expressions converge efficiently.

Alternating Series

An alternating series is a series where the terms alternate between positive and negative. This happens due to a factor like \((-1)^n\), which changes sign as \( n \) increases.

• For example, in our given series, \( \sum_{n = 1}^{\infty} \frac{(-1)^n}{\sqrt{n}} \), each next term has the opposite sign.
• Alternating series are special because they can converge even if the terms don’t get very small quickly. This is known as conditional convergence.
• To check convergence of an alternating series, we often use the Alternating Series Test, which assesses:
  • The absolute value of the terms decreases.
  • The terms approach zero as \( n \) goes to infinity.

Understanding alternating series is crucial because it extends our toolkit for dealing with series convergence. In our case, recognizing an alternating series at the endpoint \( x = \frac{1}{4} \) helped us note it converges there separately from the interval itself.