Question

Determine whether the following series converge absolutely or conditionally, or diverge. $$\sum_{k=1}^{\infty}(-1)^{k} e^{-k}$$

Short answer

Answer: The series converges absolutely.

Step-by-step solution

Absolute Convergence Test

Consider the absolute value of the series terms: $$\left|(-1)^{k} e^{-k}\right| = e^{-k}$$ Now, test the convergence of the sum of these absolute values: $$\sum_{k=1}^{\infty}e^{-k}$$

Using the Comparison Test

Since all the terms \(e^{-k}\) are positive, we can use the comparison test. Compare the series with a geometric series of the form: $$\sum_{k=1}^{\infty}r^k$$ In this case, \(r=e^{-1}\). Note that a geometric series converges if \(|r|<1\), which is true for \(e^{-1}\)

Showing the Convergence of the Geometric Series

Since \(0< e^{-1}<1\), the geometric series converges: $$\sum_{k=1}^{\infty}(e^{-1})^k = \frac{e^{-1}}{1-e^{-1}}$$

Conclude that the Series Converges Absolutely

Since our original series of absolute values is dominated by a convergent geometric series, it converges by the comparison test: $$\sum_{k=1}^{\infty}e^{-k}$$ Therefore, the original series converges absolutely. Final conclusion: The series $$\sum_{k=1}^{\infty}(-1)^{k} e^{-k}$$ converges absolutely.

Key concepts

Absolute Convergence

In the realm of infinite series, understanding the concept of absolute convergence is vital. An infinite series \( \sum a_k \) is said to converge absolutely if the series composed of its absolute values \( \sum |a_k| \) converges. This is a strong form of convergence because if a series converges absolutely, it also converges in the regular sense.

To test for absolute convergence, we take each element in the series and consider its absolute value, creating a new series. If this new series, like \( \sum_{k=1}^{\infty} e^{-k} \), converges according to standard convergence tests, then the original series \( \sum_{k=1}^{\infty}(-1)^{k} e^{-k} \) is declared to converge absolutely. This gives a clear and decisive conclusion about the behavior of the series, offering a solid ground when dealing with series that have both positive and negative terms.

Conditional Convergence

Conditional convergence arises when an infinite series \( \sum a_k \) converges, but its series of absolute values \( \sum |a_k| \) does not. In simpler terms, this means the series behaves nicely (converges) in its regular form but not when you remove the alternating nature and consider only magnitude.

A classic example of a conditionally convergent series is the alternating harmonic series. However, our given series \( \sum_{k=1}^{\infty}(-1)^{k} e^{-k} \) does not fall under this category. Since it converges absolutely, as we proved when the absolute value series converged, it cannot be conditionally convergent. Remember: absolute convergence guarantees regular convergence, but conditional convergence is usually a sign of more "delicate" balancing in the series.

Comparison Test

The comparison test is a handy tool for determining the convergence of series. It compares the series in question with another series whose convergence is already known. If \( 0 \leq a_k \leq b_k \) for all \( k \) from some index onward, then:

• If \( \sum b_k \) converges, then \( \sum a_k \) also converges.
• If \( \sum a_k \) diverges, then \( \sum b_k \) also diverges.

For our series \( \sum e^{-k} \), we used the comparison test with a geometric series, \( \sum (e^{-1})^k \), which is known to converge because \( e^{-1} < 1 \). This allowed us to conclude that \( \sum e^{-k} \) converges absolutely, thus proving absolute convergence for the original series as well.

Geometric Series

Geometric series are one of the simplest and most important types of series in mathematics. A geometric series takes the form \( \sum_{k=0}^{\infty} ar^k \), where \( a \) is the first term and \( r \) is the common ratio. The convergence of a geometric series depends on the value of \( r \):

• The series converges if \( |r| < 1 \)
• The series diverges if \( |r| \geq 1 \)

For our purposes, understanding this behavior is crucial.

In our solution, we identified that \( \sum_{k=1}^{\infty} e^{-k} \) behaves like a geometric series with ratio \( r = e^{-1} \), which is less than 1. This insight allowed us to confirm its convergence easily, as any geometric series with such \( r \) must converge. This knowledge is indispensable when facing series with exponential terms.