Problem 11 - Thomas Calculus

Question

In Exercises \(1-36\) , (a) find the series' radius and interval of convergence. For what values of \(x\) does the series converge (b) absolutely, (c) conditionally? $$ \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{n !} $$

Step-by-Step Solution

Answer

The series converges absolutely for all \( x \); no conditional convergence occurs.

Step 1: Identify the Series

The given series is a power series of the form \( \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!} \). This is a series in terms of \( x \) where each term is \( \frac{(-1)^n x^n}{n!} \).

Step 2: Find the Radius of Convergence

This series is comparable to the exponential function's power series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), which converges for all \( x \). Therefore, the radius of convergence \( R \) is \( \infty \).

Step 3: Determine the Interval of Convergence

Since the radius of convergence is \( \infty \), the interval of convergence is the entire real line, i.e., \((-\infty, \infty)\).

Step 4: Check for Absolute Convergence

For absolute convergence, consider the series \( \sum_{n=0}^{\infty} \frac{|x^n|}{n!} \), which is the same as the exponential series \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \). This series converges for all \( x \), thus it converges absolutely for all real \( x \).

Step 5: Check for Conditional Convergence

A series converges conditionally if it converges but does not converge absolutely. Since the given series converges absolutely for all \( x \), there are no values of \( x \) for which it converges conditionally.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when studying power series. It tells us how far out we can go with the variable \(x\) such that the series converges. For our specific series, which resembles the Maclaurin series for the exponential function, the radius of convergence \(R\) is \(\infty\). This means the series converges for any real number \(x\).

To find the radius of convergence, we typically apply the Ratio Test or the Root Test. For a power series \(\sum_{n=0}^{\infty} a_n x^n\), the series converges when:

• If using the Ratio Test, \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\).
• If using the Root Test, \(\lim_{n \to \infty} \sqrt[n]{|a_n|} < 1\).

In our series, the term \(\frac{x^n}{n!}\) decreases very rapidly due to the factorial in the denominator, leading to \(R = \infty\). Therefore, it's valid for all \(x\) on the real line.

Interval of Convergence

Essentially, the interval of convergence encompasses all \(x\) values where our series converges. Since for our given series, the radius of convergence is \(\infty\), the interval of convergence is \((-\infty, \infty)\).

This means every real number \(x\) makes the series converge. This infinite interval of convergence is typical for series that closely resemble well-known functions like the exponential series.

Keep in mind:

• The interval can be finite or infinite depending on the function.
• For finite intervals, endpoints often need separate testing as convergence can differ.

In our case, there is no need to check any endpoints since the interval covers the entire real line without any boundaries.

Absolute Convergence

Absolute convergence is when a series converges, regardless of the sign of each term. For a series \(\sum_{n=0}^{\infty} a_n\), it converges absolutely if \(\sum_{n=0}^{\infty} |a_n|\) also converges.

In our series, to check for absolute convergence, consider the expression \(\sum_{n=0}^{\infty} \frac{|x^n|}{n!}\). This is equivalent to the exponential function series, and since it converges for all \(x\), our original series also converges absolutely for all real \(x\).

Points to remember:

• Absolute convergence implies regular convergence.
• If a series converges absolutely, rearranging its terms won’t affect the sum.
• In cases where absolute convergence is proven, there's no conditional convergence.

In this particular example, since the series converges absolutely for every \(x\), we don’t find any values for which the series converges conditionally.