Problem 12 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} e^{-n} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} e^{-n} \) converges because it is a geometric series with a common ratio \( r = \frac{1}{e} < 1 \).

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} e^{-n} \). This is an infinite series with terms of the form \( e^{-n} \) or \( \frac{1}{e^n} \). Recognize this as a geometric series where each term is \( a_n = \left( \frac{1}{e} \right)^n \).

Step 2: Evaluate the Common Ratio

A geometric series is defined where each subsequent term is multiplied by a common ratio \( r \). Here, \( r = \frac{1}{e} \), since \( e^{-n} = \left( \frac{1}{e} \right)^n \).

Step 3: Apply Convergence Criteria for Geometric Series

For a geometric series \( \sum_{n=1}^{\infty} a r^n \) to converge, the absolute value of the common ratio must satisfy \( |r| < 1 \). Here, \( |\frac{1}{e}| < 1 \), since \( \frac{1}{e} \approx 0.368 < 1 \).

Step 4: Conclude Convergence

Since the common ratio \( \frac{1}{e} \) satisfies the condition \( |r| < 1 \), the series \( \sum_{n=1}^{\infty} e^{-n} \) converges. The series converges geometrically towards a specific sum.

Key Concepts

Convergence Criteria

When dealing with geometric series, determining whether a series converges or diverges is crucial. A series converges if its terms approach a particular value as the series extends towards infinity. For geometric series specifically, this is evaluated using the common ratio, denoted as \( r \).

The convergence criteria for geometric series states that a geometric series \( \sum_{n=0}^{\infty} ar^n \) would converge if the absolute value of the common ratio is less than one, which can be written as \( |r| < 1 \). Conversely, if \( |r| \geq 1 \), the series diverges.

In our exercise, since \( r = \frac{1}{e} \approx 0.368 \), it is clear that \( |r| < 1 \). Therefore, the given series \( \sum_{n=1}^{\infty} e^{-n} \) converges. This simple criteria is a fundamental tool in evaluating the behavior of geometric series.

Infinite Series

Infinite series refer to a sum of an infinite sequence of terms. A common way to express an infinite series is by using the formula \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the general term of the sequence.

These series can be finite (convergent) or infinite (divergent). When we say a series converges, it implies the sequence's partial sums approach a fixed number. For divergence, it means the sums increase indefinitely without reaching a limit. Solution involves verifying convergence conditions or limits.

In our exercise, recognizing \(\sum_{n=1}^{\infty} e^{-n} \) as an infinite geometric series allowed us to assess its convergence. This powerfully demonstrates how understanding series and applying convergence checks unveil mathematical insights.

Common Ratio

In geometric series, the common ratio is a key factor in defining the behavior and direction of the series. Simply put, it's the consistent factor between successive terms. If you take any term in the series, dividing it by its predecessor gives you the common ratio \( r \).

For example, in the series \( \sum_{n=0}^{\infty} ar^n \), if you start from some term \( a \) and continuously multiply by \( r \) to arrive at each next term, it forms a geometric progression.

In our particular series \( \sum_{n=1}^{\infty} e^{-n} \), each term is \( \left(\frac{1}{e}\right)^n \) with \( r = \frac{1}{e} \). Understanding the value of \( r \) (here approximately 0.368) determines if the series converges. Recognizing \( |r| < 1 \) allowed us to conclude that the series converges. The importance of determining and validating the common ratio cannot be overstated for geometric series analysis.