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Chapter 9: Problem 23

Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ \frac{x-1}{1}+\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}+\frac{(x-1)^{4}}{4}+\cdots $$

Short Answer

Expert verified
The convergence set is [0, 2).

Step by step solution

01

Identify the Series and nth-term

The given series is \( \sum_{n=1}^{\infty} \frac{(x-1)^n}{n} \). Here, each term of the series can be expressed as \( a_n = \frac{(x-1)^n}{n} \).
02

Apply the Absolute Ratio Test

The Absolute Ratio Test involves calculating the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute the expression for \( a_n \): \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-1)^{n+1}/(n+1)}{(x-1)^n/n} \right| = \left| \frac{(x-1) \cdot n}{n+1} \right|. \]
03

Simplify the Ratio

Simplify the ratio: \[ \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x-1|n}{n+1}. \] As \( n \to \infty \), \( \frac{n}{n+1} \to 1 \), therefore \[ L = \lim_{n \to \infty} \frac{|x-1|n}{n+1} = |x-1|. \]
04

Determine Convergence

For the series to converge, we require \( L < 1 \). Thus, \( |x-1| < 1 \). This implies \( -1 < x - 1 < 1 \) or \( 0 < x < 2 \).
05

Check Endpoint Convergence

Check if the series converges at\( x = 0 \) and \( x = 2 \):- At \( x = 0 \), series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \), which converges (alternating series test).- At \( x = 2 \), series becomes \( \sum_{n=1}^{\infty} \frac{1}{n} \), which diverges (harmonic series).
06

Final Step: Convergence Set

Combining the results, the set of convergence is \( 0 \leq x < 2 \).

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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 type of series where the terms involve powers of a variable, commonly denoted as \( x \). The power series takes the form:
  • \( \sum_{n=0}^{\infty} a_n (x-c)^n \) where each term is a function of \( x \).
  • The series often "centers" at a particular value \( c \), known as the center of the series.
  • For the power series given in the exercise, it centers at \( c = 1 \) because it's structured as \( \frac{(x-1)^n}{n} \).
These series are crucial in calculus and analysis because they can express complex functions in an easily integrable and differentiable form. Understanding where a power series converges, known as its convergence, is essential to using it effectively. We'll explore how to find this convergence set using various tests.
Ratio Test
The Ratio Test is a powerful tool used to determine the convergence of a series, especially of a power series. By comparing the size of successive terms, it helps us identify whether their overall sum converges to a finite value. For a series \( \sum a_n \):
  • Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If \( L < 1 \), the series absolutely converges.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive, and other methods must be considered.
In the provided exercise, the Ratio Test involves finding \( L = |x-1| \). This result tells us about how the terms behave as \( n \) increases, giving us crucial insight into where the series converges.
Convergence Set
A convergence set defines the values of \( x \) for which a power series will converge. To determine this set, we often use tests like the Ratio Test or the Alternating Series Test. Here's what you need to know:
  • From the Ratio Test in the exercise, \( |x-1| < 1 \), indicating the series converges when \( 0 < x < 2 \).
  • To fully define the convergence set, one must check endpoints separately since the Ratio Test may not apply there.
  • For \( x = 0 \) and \( x = 2 \) in our exercise, further investigation established that the series converges at \( x = 0 \) but diverges at \( x = 2 \).
So, the final convergence set given the analysis is \( 0 \leq x < 2 \), demarcating the range where the series will converge.
Alternating Series Test
The Alternating Series Test is a method to determine the convergence of series where successive terms alternate in sign, such as \( (-1)^n a_n \). The test involves a couple of simple checks:
  • The absolute value of terms \( a_n \) must be decreasing, i.e., \( a_{n+1} \le a_n \).
  • The limit of \( a_n \) as \( n \rightarrow \infty \) must be zero.
In our problem, at \( x = 0 \), the series becomes an alternating series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \). Applying the Alternating Series Test:
  • Since \( 1/n \) decreases as \( n \) increases, and the limit of \( 1/n \) is zero, the series converges at \( x = 0 \).
This test confirmed convergence at \( x = 0 \), preventing any misleading conclusions about the convergence of the entire series.

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Most popular questions from this chapter

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=x \sec \left(x^{2}\right)+\sin x\)

Let \(g(x)=p(x)+x^{n+1} f(x)\), where \(p(x)\) is a polynomial of degree at most \(n\) and \(f\) has derivatives through order \(n\) Show that \(p(x)\) is the Maclaurin polynomial of order \(n\) for \(g\).

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