Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{(x+1)^{n}}{n !} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the series

The given series is \( \sum_{n=1}^{\infty} \frac{(x+1)^{n}}{n !} \). This is a power series in the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \) with \( c_n = \frac{1}{n!} \) and \( a = -1 \). Our task is to find the convergence set for this series.

Step 2: Identify the type of series

This series is of the exponential form, specifically resembling the Taylor series expansion \( e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} \). The convergence properties of such series are known to be infinite. This means they converge for all values of \( t \). We need to investigate if our series has similar properties.

Step 3: Apply the Ratio Test

To determine the interval of convergence, apply the Ratio Test to the series: \( a_n = \frac{(x+1)^n}{n!} \). Consider \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x+1)^{n+1}}{(n+1)!} \times \frac{n!}{(x+1)^{n}} \right| = \lim_{n \to \infty} \left| \frac{x+1}{n+1} \right|. \]As \( n \to \infty \), this limit approaches 0 for all finite \( x \).

Step 4: Conclusion of the Ratio Test

Since the limit is 0, which is less than 1 for any \( x \), the Ratio Test indicates that the series converges for all finite \( x \). This means the radius of convergence is infinite.

Key Concepts

Radius of Convergence

The radius of convergence of a power series tells us how far from the center of the series, denoted by \(a\), we can move and still have the series converge. To find it, we often use the Ratio Test or Root Test. The radius of convergence, \(R\), is given by the limit:\[ R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|c_n|}}\]If \(R = \infty\), the series converges for any value of \(x\); otherwise, it converges if \(|x-a| < R\). In our exercise, since the limit of the Ratio Test gave us 0, which is less than 1 for all \(x\), this implies that \(R = \infty\). Hence, our series converges for every real number \(x\).

Ratio Test

The Ratio Test is a powerful tool for determining series convergence. It compares the ratio of successive terms of the series. For a series \(\sum_{n=0}^{\infty} a_n\), you examine:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]

• If \(L < 1\), the series converges absolutely.
• If \(L > 1\) or is infinite, the series diverges.
• If \(L = 1\), the test is inconclusive.

In our power series, because of the special form \(\frac{(x+1)^{n}}{n!}\), applying the Ratio Test gives us \(L = 0\), signifying convergence for all \(x\). This aligns with the infinite radius of convergence.

Taylor Series

Taylor series express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. The general formula for a function \(f(x)\) about a point \(a\) is:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\]This concept makes it easier to approximate functions using polynomials. For the series given in the problem, it resembles the famous Taylor series expansion for \(e^x\), specifically \(e^t\) where \(t = x+1\):\[ e^{x+1} = \sum_{n=0}^{\infty} \frac{(x+1)^n}{n!}\]This similarity is utilized to understand the convergence properties of the series.

Power Series Expansion

Power series expansions are a way to represent functions as an infinite sum of powers of \((x-a)\). A power series is generally written as:\[ \sum_{n=0}^{\infty} c_n (x-a)^n\]where \(c_n\) are coefficients, and \(x\) is a variable. Each term of the series \((c_n (x-a)^n)\) represents a polynomial expression, allowing complex functions to be expressed via simpler polynomial relationships. The series provided: \( \sum_{n=1}^{\infty} \frac{(x+1)^{n}}{n !} \) allows us to explore many properties such as convergence and approximation of functions over intervals. The expansion remains valid within its radius of convergence, enlightening how variable functions behave in a "tamed" and approachable mathematical way.