Question

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{3^{n}}{n !}\)

Short answer

The series converges absolutely.

Step-by-step solution

Identify the Series Form

The series given is \(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{3^{n}}{n !}\). This is an alternating series because of the \((-1)^{n+1}\) factor.

Check Absolute Convergence

For absolute convergence, consider the series \(\sum_{n=1}^{\infty} \left|(-1)^{n+1} \frac{3^{n}}{n!}\right| = \sum_{n=1}^{\infty} \frac{3^{n}}{n!}\). This is akin to the exponential series \( e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) for \(x=3\). Thus, it converges absolutely because the exponential series converges for all \(x\).

Conclusion on Convergence Type

Since the series \(\sum_{n=1}^{\infty} \frac{3^{n}}{n!}\) converges by the behavior of the exponential series, the absolute value of the original series also converges. Therefore, the original series converges absolutely.

Key concepts

Absolute Convergence

Absolute convergence is a crucial concept when it comes to series in calculus. It suggests that a series converges even when we take the absolute values of its terms. In simpler terms, a series \( \sum a_n \) is said to converge absolutely if \( \sum |a_n| \) is also a convergent series. Absolute convergence is particularly important because if a series converges absolutely, it is guaranteed to converge.For example, consider the series given by \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3^{n}}{n!} \). To check for absolute convergence, we need to analyze the corresponding series of absolute terms, which is \( \sum_{n=1}^{\infty} \frac{3^n}{n!} \). This is similar to an exponential series, which converges for all values of \( x \). Thus, the original series also converges absolutely.

Alternating Series

An alternating series is a series whose terms alternate in sign. These types of series typically take the form \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \), where each \( b_n \) is a positive term. The distinguishing feature is the factor \( (-1)^{n+1} \), which flips the sign of each consecutive term.To determine if an alternating series like \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3^n}{n!} \) converges, we can use the Alternating Series Test. This test requires:

• The absolute terms \( b_n \) are decreasing.
• The limit of \( b_n \) as \( n \) approaches infinity is zero.

If both conditions are satisfied, the alternating series converges.In our example, the terms \( b_n = \frac{3^n}{n!} \) decrease and approach 0 as \( n \) grows. Therefore, it passes the test, confirming that the alternating series converges.

Exponential Series

The exponential series is a powerful and foundational concept in calculus. It represents one of the simplest and most elegant infinite series: \( e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} \). Here, \( e \) is the base of the natural logarithm, approximately equal to 2.718.The exponential series converges for all values of \( x \), making it a robust tool in mathematical analysis. Each term in the series \( \frac{x^n}{n!} \) becomes smaller rapidly due to the factorial in the denominator, which ensures convergence regardless of the value of \( x \).In the problem, we identified the series \( \sum_{n=1}^{\infty} \frac{3^n}{n!} \) as a special case of the exponential series where \( x=3 \). Since the exponential series converges for any \( x \), our series also converges, confirming absolute convergence of the original alternating series in the exercise.