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Chapter 5: Problem 22

Find the radius of convergence \(R\) and interval of convergence for \(\sum a_{n} x^{n}\) with the given coefficients \(a_{n}\). $$ \sum_{n=1}^{\infty} \frac{n^{2} x^{n}}{2^{n}} $$

Short Answer

Expert verified
The radius of convergence is 2; the interval of convergence is \((-2, 2)\).

Step by step solution

01

Identify the General Formula

The given series is \( \sum_{n=1}^{\infty} \frac{n^{2} x^{n}}{2^{n}} \). Here, the coefficients \( a_n \) are given by \( a_n = \frac{n^{2}}{2^{n}} \).
02

Utilize the Ratio Test

To find the radius of convergence, use the Ratio Test. Compute the ratio \( \left| \frac{a_{n+1}}{a_{n}} \right| \) and then take the limit as \( n \to \infty \).Compute:\[\left| \frac{a_{n+1}}{a_{n}} \right| = \left| \frac{(n+1)^{2} x^{n+1}/2^{n+1}}{n^{2} x^{n}/2^{n}} \right| = \left| \frac{(n+1)^{2}}{n^{2}} \cdot \frac{x}{2} \right|.\]
03

Simplify the Ratio Expression

Simplify the expression from Step 2:\[\lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \right| \cdot \left| \frac{x}{2} \right| = \lim_{n \to \infty} \left(1 + \frac{2}{n} + \frac{1}{n^2}\right) \cdot \left(\frac{|x|}{2}\right).\]As \( n \to \infty \), the limit is \( \left| \frac{x}{2} \right| \).
04

Find the Radius of Convergence

Set the result of the Ratio Test to be less than 1 for convergence:\[ \left| \frac{x}{2} \right| < 1. \]This implies that:\[ |x| < 2.\]Thus, the radius of convergence, \( R \), is \( 2 \).
05

Determine the Interval of Convergence

The open interval of convergence from Step 4 is \((-2, 2)\). Test the endpoints by substituting \( x = 2 \) and \( x = -2 \) into the series:- For \( x = 2 \), the series becomes \( \sum \frac{n^2}{2^n} \cdot 2^n = \sum n^2 \), which diverges.- For \( x = -2 \), the series becomes \( \sum \frac{n^2 (-2)^n}{2^n} = \sum (-1)^n n^2 \), which also diverges.Thus, the series only converges in the open interval \((-2, 2)\).

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Key Concepts

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

Interval of Convergence
The interval of convergence is the range of values for which a power series converges. It's crucial to determine this interval to understand where the power series behaves well and sums up to a specific function.
In our case, we look at:
  • The result from the Ratio Test: \( |x| < 2 \)
  • From this result, we identify an open interval of convergence, which is \((-2, 2)\)
  • "Open" means that the endpoints aren't included, so we need to verify them separately
When we tested the endpoints \( x = 2 \) and \( x = -2 \), both resulted in divergent series. Hence, our interval remains open between \(-2\) and \(2\). This means the series will only converge when x is strictly between these two values.
Ratio Test
The Ratio Test is a powerful method to determine the convergence of a series. Here's how it works:
  • Take the ratio of successive terms in your series: \( \left| \frac{a_{n+1}}{a_n} \right| \)
  • Consider the limit \( n \to \infty \)
  • If the limit is less than 1, the series converges
In our example, \[\left| \frac{(n+1)^2}{n^2} \cdot \frac{x}{2} \right|\]leads to a simplified version:\[\lim_{n \to \infty} \left(1 + \frac{2}{n} + \frac{1}{n^2}\right) \cdot \left(\frac{|x|}{2}\right) = \left| \frac{x}{2} \right|\] As the limit approaches just \( |x|/2 \), setting it less than 1 provides the radius of convergence.
Power Series
A power series is a series of the form \( \sum a_n x^n \), where each term involves a power of \(x\). They resemble polynomials but with potentially infinite terms.
Here are a few key points:
  • "Power" indicates variable exponents, in this case, \(x^n\)
  • Each coefficient \(a_n\) determines the weight of each term
  • Convergence is based on the variable \(x\) sitting within a specific range
Referring back to our series, the expression \[\sum_{n=1}^{\infty} \frac{n^2 x^n}{2^n}\]is a classic example. The particular coefficients \(a_n = \frac{n^2}{2^n}\) dictate how terms like \(x^n\) grow, shaping the series convergence behavior. Understanding a power series involves deciphering these parameters and finding where \[\sum a_n x^n\]actually converges to provide meaningful results.

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Most popular questions from this chapter

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of f. $$ f(x)=e^{x} \cos x $$

In the following exercises, find the Taylor series at the given value. $$ f(x)=\frac{3}{x}, a=1 $$

In the following exercises, find the radius of convergence of the Maclaurin series of each function. $$ \frac{1}{1+x^{2}} $$

In the following exercises, compute the Taylor series of each function around \(x=1\). $$ f(x)=\ln x $$

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of f. $$ f(x)=\ln (\cos x) $$
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