Problem 45 - Calculus: Early Transcendentals

Question

(a) Show that \( \sum_{n=0}^{\infty} x^n/n! \) converges for all \( x. \) (b) Deduce that \( lim_{n \to \infty} x^n/n! = 0 \) for all \( x. \)

Step-by-Step Solution

Answer

(a) The series converges for all \( x \). (b) \( \lim_{n \to \infty} \frac{x^n}{n!} = 0 \) for all \( x \).

Step 1: Identify the Series

The series in question is the exponential series given by \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \). This is a power series in \( x \).

Step 2: Apply the Ratio Test

To determine the convergence of the series, we use the Ratio Test. For the series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), consider the ratio \( a_{n+1}/a_n = \frac{x^{n+1}}{(n+1)!} \bigg/ \frac{x^n}{n!} = \frac{x}{n+1} \).

Step 3: Evaluate the Limit in the Ratio Test

Compute the limit: \( \lim_{n \to \infty} \left| \frac{x}{n+1} \right| = \lim_{n \to \infty} \frac{|x|}{n+1} = 0 \). The limit is 0 for any finite \( x \), so the Ratio Test indicates that the series converges for all \( x \).

Step 4: Conceptual Deduction of Limit on Series Terms

For part (b), notice that convergence of series implies that its general term goes to zero. Specifically, if a series \( \sum a_n \) converges, then \( \lim_{n \to \infty} a_n = 0 \).

Step 5: Confirming the Limit

Since \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) converges for any \( x \), the term \( \frac{x^n}{n!} \) must approach zero as \( n \to \infty \). Thus, \( \lim_{n \to \infty} \frac{x^n}{n!} = 0 \) for any \( x \).

Key Concepts

Ratio Test

The Ratio Test is a powerful tool for determining the convergence of a series. To use this test, we examine the series' ratio of successive terms. For a series \( \sum a_n \), consider the ratio \( \frac{a_{n+1}}{a_n} \). If the limit of this ratio as \( n \to \infty \) is less than 1, the series converges absolutely.

In our exercise, we apply the Ratio Test to the exponential series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \). We compute the ratio of successive terms: \( \frac{x^{n+1}}{(n+1)!} \bigg/ \frac{x^n}{n!} = \frac{x}{n+1} \). The next step is to calculate the limit of this result as \( n \to \infty \), which leads us to \( \lim_{n \to \infty} \frac{|x|}{n+1} = 0 \).

• This result shows that, regardless of the value of \( x \), the limit is always 0, indicating convergence for any \( x \).

Exponential Series

Exponential series are particularly interesting in mathematics because they generalize the exponential function. The series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) is known as the exponential series. It's one of the simplest and most significant power series.

An important property of exponential series is that they converge for all real numbers \( x \). This result is central to calculus and helps in understanding functions beyond polynomials.

• The exponential series connects to the familiar exponential function \( e^x \), which can be expressed by this series.
• This series' convergence for any \( x \) comes from the rapid growth of the factorial in the denominator.

By multiplying \( x^n \) by the rapidly increasing \( n! \) in the denominator, the terms of the series shrink sufficiently to guarantee convergence.

Convergence

Understanding convergence is crucial for analyzing series and functions. A series converges if the sequence of its partial sums approaches a specific finite value as more terms are added. In other words, the series \( \sum_{n=0}^{\infty} a_n \) converges if there exists a finite limit to \( S_N = a_0 + a_1 + \cdots + a_N \) as \( N \to \infty \).

For our exponential series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), we have shown it converges for all \( x \) by using the Ratio Test. This result means that if you continue adding more and more terms of the series, they will sum up to a well-defined, finite number.

• This convergence implies that series’ terms decrease at a rate that ensures no infinite value or oscillation.
• The convergence of a series is essential in calculus for defining and working with functions in a broad range of applications.

Limits

Limits are foundational concepts in calculus, providing a way to analyze the behavior of functions as inputs approach a certain value. In series, limits help us understand the long-term behavior of sequences.

Applied to our exercise, the concept of limits shows us that for all \( x \), \( \lim_{n \to \infty} \frac{x^n}{n!} = 0 \). This limit is derived from the convergence of the exponential series, indicating that each term becomes negligible as \( n \to \infty \).

• If the terms of a series don't tend towards zero, the series cannot converge.
• Thus, demonstrating that these terms do go to zero is key to proving convergence.

Limits are not only essential for establishing convergence but also valuable in other areas of calculus, particularly when working with derivatives and integrals.