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Magazine Chapter 9: Problem 30
Find the interval of convergence. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}(x-5)^{n}}{2^{n} n^{2}}$$
Short Answer
Expert verified
The interval of convergence is \([3, 7]\).
Step by step solution
01
Identify the Power Series
The given series is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(x-5)^{n}}{2^{n} n^{2}} \). This is a power series centered at \( x = 5 \) with terms \( a_n = \frac{(-1)^{n}(x-5)^{n}}{2^{n} n^{2}} \).
02
Apply the Ratio Test
Use the Ratio Test to find the radius of convergence. Start by computing the ratio: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute \( a_n \) into the formula: \[ \lim_{n \to \infty} \left| \frac{(-1)^{n+1}(x-5)^{n+1}}{2^{n+1} (n+1)^2} \times \frac{2^n n^2}{(-1)^n (x-5)^n} \right| \].
03
Simplify the Ratio
Simplify the expression from Step 2 to get: \[ \lim_{n \to \infty} \left| \frac{(x-5)}{2} \times \frac{n^2}{(n+1)^2} \right| = \left| \frac{x-5}{2} \right| \cdot \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^2 \].
04
Evaluate the Limit
Evaluate the remaining limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^2 = 1 \].
05
Find the Radius of Convergence
The ratio from the Ratio Test gives: \[ \left| \frac{x-5}{2} \right| < 1 \]. Thus, \( \frac{|x-5|}{2} < 1 \) implies \( |x-5| < 2 \), which means \( R = 2 \).
06
Determine the Interval of Convergence
The inequality from Step 5 gives \( 3 < x < 7 \). Now check the endpoints \( x = 3 \) and \( x = 7 \) for convergence.
07
Test Endpoint \( x = 3 \)
Substitute \( x = 3 \): The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(-2)^{n}}{2^{n} n^{2}} = \sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n^{2}} = \sum_{n=1}^{\infty} \frac{1}{n^{2}} \), which converges by the p-series test (p=2>1).
08
Test Endpoint \( x = 7 \)
Substitute \( x = 7 \): The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(2)^{n}}{2^{n} n^{2}} = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \), which converges by the alternating series test.
09
State the Final Interval of Convergence
Since both endpoints \( x = 3 \) and \( x = 7 \) result in convergent series, the interval of convergence is \[ x \in [3, 7] \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is like a super-charged polynomial that can have infinitely many terms. You might see them written as \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where each term \( a_n (x-c)^n \) can be thought of as a part of the polynomial that stretches on forever. Here, \( c \) is the center of the series, and the series converges or diverges depending on the values of \( x \) you choose.
- Center: The point \( x = c \) around which the series is centered.
- Terms: Each term contains \( (x-c)^n \), which highlights how the series 'centers' around \( c \).
- Convergence: A power series doesn鈥檛 converge everywhere; understanding where it does helps us with problems like finding its interval of convergence.
Ratio Test
The Ratio Test is a handy tool for determining if a series converges. Here's how it works:
- Compute the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
- If this limit is less than 1, the series converges.
- If the limit is greater than 1, the series diverges. If it equals 1, the test is inconclusive.
Radius of Convergence
The Radius of Convergence, symbolized as \( R \), tells us how far across the \( x \)-axis our power series radiates from its center. It鈥檚 the half-width of the interval within which the series converges.
- Formula: Often derived from the Ratio Test results \( \left| \frac{x-c}{R} \right| < 1 \).
- Interval Relation: This translates into \( c - R < x < c + R \).
Alternating Series Test
An alternating series is a special type of series where signs flip between positive and negative. The Alternating Series Test helps determine whether such a series converges.
- Two conditions must be checked:
- The absolute value of the terms is decreasing.
- The limit of the terms as \( n \to \infty \) is zero.
