Math Problem Statement

what's the radius and interval of convergence of the series ((-1)^n(x-2)^n)/(n2^n)

Solution

To determine the radius and interval of convergence of the series

n=1(1)n(x2)nn2n\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n 2^n}

we will use the Ratio Test to find the radius of convergence, and then identify the interval.

Step 1: Apply the Ratio Test

The Ratio Test is used to find the radius of convergence for power series. The general form of the Ratio Test is:

L=limnan+1anL = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

For the given series, the general term ana_n is:

an=(1)n(x2)nn2na_n = \frac{(-1)^n (x-2)^n}{n 2^n}

Now, apply the Ratio Test:

an+1an=(1)n+1(x2)n+1(n+1)2n+1×n2n(1)n(x2)n\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} (x-2)^{n+1}}{(n+1) 2^{n+1}} \times \frac{n 2^n}{(-1)^n (x-2)^n}

Simplifying:

an+1an=(1)n+1(x2)n+1n2n(1)n(x2)n(n+1)2n+1=(x2)2×nn+1\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} (x-2)^{n+1} n 2^n}{(-1)^n (x-2)^n (n+1) 2^{n+1}} = \frac{(x-2)}{2} \times \frac{n}{n+1}

Now, take the limit as nn \to \infty:

limnan+1an=x22\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x-2}{2} \right|

For the series to converge, we require this limit to be less than 1:

x22<1\left| \frac{x-2}{2} \right| < 1

Step 2: Solve for the Radius of Convergence

Solve the inequality:

x22<1\left| \frac{x-2}{2} \right| < 1

Multiplying both sides by 2:

x2<2|x - 2| < 2

This gives the radius of convergence R=2R = 2, and the series converges for xx in the interval:

(22,2+2)=(0,4)(2 - 2, 2 + 2) = (0, 4)

Step 3: Check the Endpoints

We need to check whether the series converges at the endpoints x=0x = 0 and x=4x = 4.

  1. At x=0x = 0: The series becomes:

n=1(1)n(2)nn2n=n=1(1)n(1)nn=n=11n\sum_{n=1}^{\infty} \frac{(-1)^n (-2)^n}{n 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n (-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}

This is the harmonic series, which diverges.

  1. At x=4x = 4: The series becomes:

n=1(1)n(2)nn2n=n=1(1)nn\sum_{n=1}^{\infty} \frac{(-1)^n (2)^n}{n 2^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}

This is the alternating harmonic series, which converges.

Conclusion:

  • Radius of Convergence: R=2R = 2
  • Interval of Convergence: 0<x40 < x \leq 4, or (0,4](0, 4]

Would you like further details on any of the steps? Here are some related questions:

  1. How does the Ratio Test work in determining the radius of convergence?
  2. What happens if we apply the Root Test instead of the Ratio Test?
  3. Why does the alternating harmonic series converge, and what tests can confirm this?
  4. How can we graphically represent the convergence of power series?
  5. What is the difference between conditional and absolute convergence?

Tip: Always check the behavior at the endpoints of the interval when using the Ratio Test to determine convergence!

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Math Problem Analysis

Mathematical Concepts

Power Series
Convergence
Ratio Test
Harmonic Series
Alternating Series

Formulas

Ratio Test: L = lim(n -> ∞) |a_(n+1)/a_n|
|(x - 2) / 2| < 1

Theorems

Ratio Test
Harmonic Series Convergence
Alternating Series Test

Suitable Grade Level

College/Advanced High School (Grades 11-12, Undergraduate)