/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Find the interval of convergence... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the interval of convergence of each power series. \(\sum_{n=1}^{x} \frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n}\)

Short Answer

Expert verified
The interval of convergence is \([-8, 12)\).

Step by step solution

01

Initial Setup and Identification

First, let's identify the given power series: \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n} \). This is a power series centered at \(x = 2\). Our goal is to find the interval of convergence of this series.
02

Apply the Ratio Test

To find the interval of convergence, we use the Ratio Test. Consider \(a_n = \frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n}\). The Ratio Test requires evaluating \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+2}}{(n+1) \cdot 10^{n+1}}(x-2)^{n+1}}{\frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n}} \right|.\]
03

Simplify the Ratio

Simplifying the expression from the Ratio Test:\[\left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{n}{n+1} \cdot \frac{(x-2)}{10} \right| = \lim_{n \to \infty} \frac{n}{n+1} \cdot \left| \frac{x-2}{10} \right|.\]As \(n\) approaches infinity, \( \frac{n}{n+1} \to 1 \). So, the limit becomes:\[\left| \frac{x-2}{10} \right| < 1.\]
04

Solve the Inequality for Convergence

Solve the inequality \( \left| \frac{x-2}{10} \right| < 1\):\[\left| x-2 \right| < 10.\]This can be rewritten as:\[-10 < x-2 < 10.\]Adding 2 to each part, we get:\[-8 < x < 12.\]
05

Testing Endpoints

Check the endpoints \(x = -8\) and \(x = 12\) for convergence by substituting them back into the original series.1. At \(x = -8\): The series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cdot (-1)^n \), which converges (alternating harmonic series).2. At \(x = 12\): The series becomes \(\sum_{n=1}^{\infty} \frac{1}{n} \), which diverges (harmonic series).
06

Determine the Interval of Convergence

Since the series converges at \(x = -8\) but not at \(x = 12\), the interval of convergence is \([-8, 12)\).

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Key Concepts

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

Power Series
Power series are expressions of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \). These series are centered at a specific point \(c\), and involve terms that become increasingly more complex as \(n\) increases.
In the provided exercise, we are dealing with a power series centered at \(x = 2\), written as:
  • \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n} \)
The coefficients, \(a_n\), in this series are \( \frac{(-1)^{n+1}}{n \cdot 10^{n}} \), and the variable part is expressed as \((x-2)^n\). These elements are crucial in determining the series’ behavior, particularly in terms of convergence and where it converges.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of a series. It simplifies and quantifies how the terms of a series grow relative to each other.
To apply the Ratio Test, you analyze the series’ terms, \(a_n\), and calculate the \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This gives a clear criterion:
  • If the limit is less than 1, the series converges absolutely.
  • If the limit is greater than 1, the series diverges.
  • When the limit equals 1, the test is inconclusive.
In our example, we focused on the terms \(a_n = \frac{(-1)^{n+1}}{n \cdot 10^{n}}(x-2)^{n}\) and found:\[ \left| \frac{x-2}{10} \right| < 1. \]This inequality indicates the possible range of \(x\) where the series converges, setting up for further steps.
Convergence of Series
Understanding convergence means knowing when a series sums up to a finite value. For our power series, this convergence depends on the values of \(x\) in relation to the series' behavior. The derived inequality from the ratio test, \( \left| x-2 \right| < 10 \), translates into the interval \(-8 < x < 12\). This is where the series has the potential to converge.
Add 2 to each term of the inequality to account for the center of the series, ensuring it focuses on its convergence interval related to \(x\).When a series converges, it means the infinite sum of its terms approaches a specific, finite value. This behavior is tested at the interval's endpoints to check whether they are included.
Endpoints Testing
Endpoints testing involves checking the convergence of a series at the borders of its interval, determined through the ratio test.In the solution:
  • We tested \(x = -8\) by substituting into the original series, leading to an alternating harmonic series which converges.
  • We also tested \(x = 12\), yielding a harmonic series which is known to diverge.
The alternating harmonic series converges because its terms illustrate alternating signs and decreasing magnitude, a characteristic known for converging. On the contrary, the harmonic series diverges due to its terms diminishing too slowly to sum to a finite value.Thus, the series converges at \(x = -8\) but not at \(x = 12\), concluding the interval of convergence to be \([-8, 12)\). This valuable step finalizes the exact range where our power series remains bounded and convergent.

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