Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The series converges for all real numbers \( x \).

Step 1: Determine the General Term of the Series

The given series is a power series. Recognizing that this resembles the Taylor series expansion for the exponential function, we can identify the general term. The general term, \( a_n \), of this series is given by \( a_n = \frac{(x+2)^n}{n!} \).

Step 2: Apply the Absolute Ratio Test

The Absolute Ratio Test states that a series \( \sum a_n \) converges if the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \).Substitute the general terms to find:\[ \frac{a_{n+1}}{a_n} = \frac{\frac{(x+2)^{n+1}}{(n+1)!}}{\frac{(x+2)^n}{n!}} = \frac{(x+2)^{n+1}}{(n+1)!} \times \frac{n!}{(x+2)^n} \]Simplifying gives:\[ \frac{a_{n+1}}{a_n} = \frac{(x+2)}{n+1} \]Evaluate the limit:\[ L = \lim_{n \to \infty} \left| \frac{(x+2)}{n+1} \right| = 0 \] (since the numerator is constant and the denominator approaches infinity).

Step 3: Determine the Convergence Set

Since \( L = 0 < 1 \) for all \( x \), the series converges for all values of \( x \). Therefore, the convergence set of the series is the entire real line.

Key Concepts

Absolute Ratio Test

In order to determine if a power series converges, one effective method is the Absolute Ratio Test. This test helps us to understand when a series is behaving nicely, that is, when it adds up to a finite number. For any given series, the test looks at the limit of the absolute value of the ratio between consecutive terms.
This is expressed with the formula: \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \]

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive; other methods must be used instead.

In the example from the original exercise, we find this limit and see that it equals 0, showing convergence. It’s important to note that even though the numerator \( x + 2 \) may change as \( x \) varies, in this particular example the limit simplifies to 0 regardless, due to the denominator \( n+1 \) going to infinity.

Taylor series expansion

Taylor series expansion is a way to represent a function as an infinite sum of terms that are calculated based on the function's derivatives at a single point. This is especially useful for functions that are very complicated or not easily computable. Each term in a Taylor series introduces an additional layer of approximation to the target function.
The general form of a Taylor series expansion about an input \( a \) is: \[ f(x) = \sum_{n = 0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] For the series provided in the original exercise, this corresponds to the expansion of the exponential function \( e^{x} \) about the point \( -2 \).

• Each term becomes \( \frac{(x+2)^n}{n!} \), which aligns perfectly with the series given.
• The point \( -2 \) is known as the 'center' of the series.
• The Taylor series offers a close approximation for functions around this center point, and becomes exact as an infinite series.

General Term of a Series

Recognizing the general term of a series is crucial for understanding its behavior in terms of convergence and determining the sum. The general term essentially gives the rule for forming each term in the series. For power series, which have terms of the form \( a_n(x-c)^n \), finding the general term allows us to apply tests like the Ratio Test to check for convergence.
For the series in our exercise, the general term \( a_n \) was identified as: \[ a_n = \frac{(x+2)^n}{n!} \]

• The numerator \((x+2)^n\) indicates the series powers of \( x \) are shifted by 2.
• The denominator \(n!\) results in factorial growth, which tends to make the series converge faster because factorial terms increase rapidly.
• Identifying the general term is key for applying other mathematical analysis techniques.

This deeper understanding of the general term helps in applying tests, and can also give insights into the radius and interval of convergence of the series, though that was quickly resolved in this example due to the behavior of factorial terms.