Problem 12 Find the interval of convergence... [FREE SOLUTION]

Chapter 1: Problem 12

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} n(-2 x)^{n}$$

Short Answer

Expert verified

The interval of convergence is $( -\frac{1}{2}, \frac{1}{2} )$.

Step by step solution

Determine the radius of convergence using the Ratio Test

Apply the Ratio Test to the power series $\sum_{n=1}^{\infty} n(-2x)^{n}$. The Ratio Test involves evaluating $\lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right|$, where $a_n = n(-2x)^n$. Calculate this limit.

Apply the Ratio Test

First, write $a_n = n(-2x)^n$ and $a_{n+1} = (n+1)(-2x)^{n+1}$. Then, find the limit:\[\lim_{{n \to \infty}} \left| \frac{(n+1)(-2x)^{n+1}}{n(-2x)^n} \right| = \lim_{{n \to \infty}} \left| \frac{(n+1)(-2x) \cdot (-2x)^n}{n \cdot (-2x)^n} \right| = \lim_{{n \to \infty}} \left| (n+1) \cdot \frac{-2x}{n} \right| = \lim_{{n \to \infty}} \frac{(n+1)}{n} \cdot |-2x| = 2|x|. \]

Set up the interval for convergence

For the series to converge, the limit found in Step 2 must be less than 1:\[2|x| < 1\].Therefore, solve this inequality for $x$ to find the interval of convergence:\[|x| < \frac{1}{2}\], so the interval is $ -\frac{1}{2} < x < \frac{1}{2} $.

Test the endpoints of the interval

Test the endpoints $x = -\frac{1}{2}$ and $x = \frac{1}{2}$. Plug these values into the original series and determine whether the series converges at these points.

Evaluate the series at the endpoints

For $x = \frac{1}{2}$:\[\sum_{n=1}^{\infty} n \left(-2 \cdot \frac{1}{2}\right)^n = \sum_{n=1}^{\infty} n (-1)^n. \]This is a divergent series because it does not approach zero.For $x = -\frac{1}{2}$:\[\sum_{n=1}^{\infty} n \left(-2 \cdot -\frac{1}{2}\right)^n = \sum_{n=1}^{\infty} n (1)^n. \]This is also a divergent series because it grows without bound. Therefore, the endpoints are not included in the interval of convergence.

State the final interval of convergence

Since the series converges only within the interval and not at the endpoints, the interval of convergence is $( -\frac{1}{2}, \frac{1}{2} )$.

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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 a series of the form $\textstyle \sum_{n=0}^{\infty} a_n (x - c)^n $ where $a_n$ are coefficients and $c$ is the center of the series. In simpler terms, it is a way to represent functions as an infinite sum of terms. Each term involves a power of the variable $x$ and a coefficient. Power series are very powerful because they can be used to represent a wide variety of functions and can be manipulated much like polynomials. Understanding the convergence of a power series is crucial to using it effectively. We need to determine where the series converges, which leads us to concepts like the Ratio Test and radius of convergence.

Ratio Test

The Ratio Test is a method for determining the convergence or divergence of an infinite series. Specifically, for a series $\textstyle \sum a_n $, the Ratio Test examines the limit: \[\textstyle L = \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| \]. If $ L < 1$, the series converges. If $ L > 1$, the series diverges. If $L = 1$, the test is inconclusive.

For the power series $\textstyle \sum_{n=1}^{\infty} n(-2 x)^n $, applying the Ratio Test involves calculating $\textstyle \lim_{{n \to \infty}} \left| \frac{a_{n+1}}{a_n} \right| $. This helps us find the range of $x$ values for which the series converges.

radius of convergence

The radius of convergence is the distance from the center $c$ within which a power series converges. It can be found using the Ratio Test. For the series $\textstyle \sum_{n=1}^{\infty} n(-2 x)^n $, we found $ L = 2|x| $ and set $ 2|x| < 1$. Solving this inequality gives us $ |x| < \frac{1}{2} $, indicating the series converges for $ |x| \lt \frac{1}{2} $.

This means the radius of convergence is $ \frac{1}{2} $. Inside this interval, we can be confident the series converges.

endpoint analysis

After finding the radius of convergence, it's also necessary to check whether the series converges at the endpoints of the interval. For the series $\textstyle \sum_{n=1}^{\infty} n(-2 x)^n $ with the interval $ -\frac{1}{2} < x < \frac{1}{2} $, we need to test $ x = -\frac{1}{2} $ and $ x = \frac{1}{2} $.

For $ x = \frac{1}{2} $, the series becomes $\textstyle \sum_{n=1}^{\infty} n (-1)^n $ which diverges. For $ x = -\frac{1}{2} $, the series becomes $\textstyle \sum_{n=1}^{\infty} n (1)^n $, which also diverges.
Therefore, endpoints are not included, and the final interval of convergence is $ ( -\frac{1}{2}, \frac{1}{2} ) $.

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