Question

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

Short answer

The series converges absolutely.

Step-by-step solution

Understand the Series

The given series is \[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n!}.\]The series is alternating because of the factor \((-1)^{n+1}\), which changes sign with each term.

Check for Absolute Convergence

To determine if the series converges absolutely, we consider the series of absolute values:\[\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{n!} \right| = \sum_{n=1}^{\infty} \frac{1}{n!}.\]The series \(\sum_{n=1}^{\infty} \frac{1}{n!}\) is a well-known convergent series called the exponential series for \(e\), since \(e = \sum_{n=0}^{\infty} \frac{1}{n!}.\) Thus, this series converges absolutely.

Conclude on Type of Convergence

Since the series \(\sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{n!} \right|\) converges, the original series \(\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n!}\) converges absolutely.

Key concepts

Alternating Series

An alternating series is a series where the signs of its terms alternate between positive and negative. Such series have the form \( a_1 - a_2 + a_3 - a_4 + \ldots \) or more generally, \[ \sum_{n=1}^{\infty} (-1)^n a_n, \]where the terms alternate in sign.

• In our series, \( (-1)^{n+1} \cdot \frac{1}{n!} \), the \( (-1)^{n+1} \) term is responsible for the alternating signs.
• This type of series can be tricky because the series might be converging differently depending on whether you consider the absolute values of the terms.

To determine if an alternating series converges, it's often useful to apply the Alternating Series Test. The conditions are:

• The absolute values of the terms \( a_n \) should be decreasing.
• The limit of \( a_n \) as \( n \) approaches infinity should be zero.

If both conditions are met, the alternating series is guaranteed to converge. However, it doesn’t tell us whether the convergence is absolute or conditional.

Absolute Convergence

Absolute convergence is a stronger form of convergence for a series. If a series absolutely converges, it means that when you take the absolute value of each term and sum them up, the series still converges. Mathematically, if a series \( \sum a_n \) is such that \( \sum |a_n| \) converges, then we say \( \sum a_n \) converges absolutely.

• In simpler terms, removing the signs (whether positive or negative) from each term does not affect the overall convergence of the series.
• For our series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n!} \), we considered its absolute version, \( \sum_{n=1}^{\infty} \frac{1}{n!} \), which is known to converge, hence the original series converges absolutely.

Absolute convergence is important because if a series converges absolutely, then the order of the terms can be rearranged without affecting the sum. This is different from conditional convergence where rearranging terms can potentially change the sum.

Exponential Series

The exponential series is a special type of series that is extensively used in mathematics. The general form of an exponential series is given by:\[ \sum_{n=0}^{\infty} \frac{x^n}{n!}, \]which converges for all real values of \( x \).

• A specific case of this series is when \( x = 1 \). Here, we get the series \( \sum_{n=0}^{\infty} \frac{1}{n!} \), which converges to \( e \), the base of the natural logarithm.
• It's a very well-behaved series and is known to converge to the number \( e \) quite quickly despite the infinite terms.

In the context of our exercise, we derived that the series \( \sum_{n=1}^{\infty} \frac{1}{n!} \) is indeed an exponential series, confirming its absolute convergence. Recognizing the exponential series is beneficial because it reveals how certain infinite series approach finite values, illustrating the fascinating links between infinite processes and finite numbers.