Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9 !}-\cdots $$

Step-by-Step Solution

Answer

The series converges for all real values of \(x\).

Step 1: Identify the General Term of the Series

The given power series can be written as an alternating series. Observing the pattern of coefficients and exponents, the nth term of the series is given by \((-1)^n \frac{x^{2n+1}}{(2n+1)!}\). This accounts for each term transitioning from odd powers of \(x\): \(x, x^3, x^5,\) and so on.

Step 2: Apply the Absolute Ratio Test

To use the Absolute Ratio Test, we start with the nth term \(a_n = (-1)^n \frac{x^{2n+1}}{(2n+1)!}\). The absolute value is \( |a_n| = \frac{x^{2n+1}}{(2n+1)!} \). Calculate the ratio of consecutive terms: \[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{2(n+1)+1}}{(2(n+1)+1)!} \cdot \frac{(2n+1)!}{x^{2n+1}} \right| = \left| \frac{x^2}{(2n+3)(2n+2)} \right|\]

Step 3: Determine the Limit of the Ratio as \(n \to \infty\)

Simplifying the expression from Step 2, we focus on \[\lim_{n \to \infty} \left| \frac{x^2}{(2n+3)(2n+2)} \right| = 0,\]which goes to zero for any real value of \(x\). Since this limit is always zero, the series converges for all real numbers \(x\).

Key Concepts

Absolute Ratio Test

The Absolute Ratio Test is a method used in calculus to determine the convergence or divergence of infinite series. It involves calculating the ratio of the absolute value of consecutive terms in a series. For this test to be applicable, each term must be nonzero.

In practice, the Absolute Ratio Test examines the limit:

• Given a series with terms \(a_n\), compute \( \left| \frac{a_{n+1}}{a_n} \right| \).
• Determine the limit of this ratio as \(n\), the term number, tends to infinity: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

If this limit is less than 1, the series converges absolutely. If it is more than 1, or doesn't exist, the series diverges. In our problem, we found this limit was 0, implying convergence for any real \(x\).

Convergence

Convergence in the context of a series, describes a situation where the series approaches a finite value as the number of terms increases.The series from the exercise converges if, as you add more and more terms, the total sum approaches a specific and finite number.

There are different types of convergence:

• Absolute Convergence: A series is absolutely convergent if the series of absolute values of its terms is convergent. This is usually a strong indication of overall convergence.
• Conditional Convergence: A series converges conditionally if it converges, but not absolutely. That means the series itself adds up to a finite value, but the series of absolute values does not.

For the given power series in the exercise, since the result of the Absolute Ratio Test was less than 1 for any real value of \(x\), the series converges absolutely.

Alternating Series

An alternating series is a series in which the signs of the series terms alternate between positive and negative. This type of series often appears in mathematics, especially in Fourier series and Taylor series expansions.In the exercise, the power series is expressed as an alternating series due to the factor \((-1)^n\), which changes the sign of each subsequent term:

• The first term is positive: \(x\).
• The second term is negative: \(-\frac{x^3}{3!}\).
• The third term returns to positive: \(+\frac{x^5}{5!}\), and so on.

Alternating series have their own convergence criteria known as the Alternating Series Test, which states that if the absolute values of the terms decrease to zero, and the terms are decreasing in magnitude, the series converges.

General Term

The general term in a series is a formula that allows us to represent any term based on its position in the sequence, typically denoted by \(n\).For the series given in the exercise, the general term formula was identified as:\[a_n = (-1)^n \frac{x^{2n+1}}{(2n+1)!}\]This formula tells us:

• The alternating sign is represented by \((-1)^n\), ensuring the series is alternating.
• The power of \(x\) follows an odd pattern \(x, x^3, x^5, ...\), which is reflected in \(x^{2n+1}\).
• The factorial \((2n+1)!\) is used to correctly match the denominator of each term.

Understanding the general term is crucial, as it defines how each term in the series relates to the others, providing a foundation for further analysis, such as applying convergence tests.