Question

Is the series convergent or divergent? If convergent, is it absolutely convergent? $$ \sum_{n=1}^{\infty} \cos (\pi n) e^{-n} $$

Short answer

The series is absolutely convergent.

Step-by-step solution

Analyze Series Terms

Consider the terms of the given series: \( \cos(\pi n) e^{-n} \). Notice that \( \cos(\pi n) \) equals \(-1\) for odd \(n\) and \(1\) for even \(n\). Thus, the series becomes an alternating series: \( \sum_{n=1}^{\infty} (-1)^n e^{-n} \).

Check for Convergence using Alternating Series Test

To use the Alternating Series Test, check that the magnitude of the terms \( e^{-n} \) decreases to 0 and that \( e^{-n} \) is positive. Both of these conditions hold since \( e^{-n} \rightarrow 0 \) as \( n \rightarrow \infty \) and \( e^{-n} > 0 \) for all \( n \). Therefore, the series converges by the Alternating Series Test.

Test for Absolute Convergence

Check the absolute convergence by considering the series \( \sum_{n=1}^{\infty} |\cos(\pi n) e^{-n}| = \sum_{n=1}^{\infty} e^{-n} \). This is a geometric series with a common ratio \( r = e^{-1} < 1 \). Since all terms are positive and the series converges, the original series is absolutely convergent.

Conclusion

The series \( \sum_{n=1}^{\infty} \cos(\pi n) e^{-n} \) is absolutely convergent because the absolute series \( \sum_{n=1}^{\infty} e^{-n} \) converges.

Key concepts

Alternating Series Test

The Alternating Series Test helps us determine the convergence of a series where the terms alternate in sign. Let's break it down! An alternating series has terms that flip between positive and negative. For example, a simple alternating series might look like \( a_1 - a_2 + a_3 - a_4 + ... \).
The Alternating Series Test says that if you have such a series \(\sum (-1)^n a_n \), it will converge if two conditions are met:

• The absolute value of the terms \( a_n \) should eventually decrease, meaning \( a_{n+1} \leq a_n \) for all \( n \) past a certain point.
• The terms \( a_n \) should approach zero as \( n \) becomes very large, \( \lim_{n \to \infty} a_n = 0 \).

In our exercise, the series \( \sum (-1)^n e^{-n} \) satisfies both these because \( e^{-n} \) becomes smaller as \( n \) gets larger, and it always remains positive. Thus, the Alternating Series Test confirms that it converges.

Absolute Convergence

Absolute convergence is a stronger form of convergence. To explain it simply—if a series converges absolutely, it means that the series \( \sum |a_n| \) (the sum of the absolute values of the original series) also converges.
Imagine the series \( \sum (-1)^n a_n \). It converges absolutely if \( \sum |a_n| \) converges. This concept is crucial because it assures that the series converges no matter how the terms are ordered.
In the given exercise, we examine the series \( \sum e^{-n} \), which is the absolute version of our original alternating series. This is a geometric series with the common ratio of \( e^{-1} \), which is less than 1. Since the series \( \sum e^{-n} \) converges, we conclude that the original alternating series is absolutely convergent as well.

Geometric Series

Geometric series are some of the simplest types of series to comprehend. The shape of a geometric series is quite regular—each term is multiplied by a constant factor, called the common ratio. This creates a pattern like:\( a, ar, ar^2, ar^3, ... \)
Here, \( a \) is the first term, and \( r \) is the constant ratio between terms. A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges when the absolute value of the common ratio \( |r| \) is less than 1. The sum of a convergent geometric series is given by:\[ S = \frac{a}{1 - r} \]
In the problem's solution, we notice \( \sum e^{-n} \) forms a geometric series because each term is \( e^{-n} \), translating to \( a = 1 \) and \( r = e^{-1} \). Since \( e^{-1} < 1 \), this series converges, indicating absolute convergence for the alternating series. Geometric series provide a straightforward way to confirm both convergence and absolute convergence.