Problem 12 - Calculus: Early Transcendentals

Question

Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{k = 1}^{\infty} ke^{- k} \)

Step-by-Step Solution

Answer

The series is convergent by the Ratio Test.

Step 1: Identify the terms

Identify the general term of the series. Here, the general term of the series is given by \( a_k = k e^{-k} \).

Step 2: Apply the Ratio Test Formula

The Ratio Test states that for the series \( \sum a_k \), compute the limit \( L = \lim_{{k \to \infty}} \left| \frac{a_{k+1}}{a_k} \right| \). If \( L < 1 \), the series converges, if \( L > 1 \) or \( L = \infty \), the series diverges, and if \( L = 1 \), the test is inconclusive.

Step 3: Compute \( a_{k+1} \)

Calculate \( a_{k+1} \) by substituting \( k+1 \) into the general term: \( a_{k+1} = (k+1)e^{-(k+1)} = (k+1)e^{-k}e^{-1} \).

Step 4: Set up the Ratio

Determine the ratio \( \frac{a_{k+1}}{a_k} \):\[ \frac{a_{k+1}}{a_k} = \frac{(k+1)e^{-k}e^{-1}}{ke^{-k}} = \frac{k+1}{k} \cdot e^{-1}. \]

Step 5: Simplify the ratio

Simplify \( \frac{k+1}{k} \): \( \frac{k+1}{k} = 1 + \frac{1}{k} \).Thus, the ratio becomes:\[ \frac{a_{k+1}}{a_k} = \left(1 + \frac{1}{k}\right) e^{-1}. \]

Step 6: Calculate the Limit

Take the limit as \( k \to \infty \) of the ratio:\[ L = \lim_{{k \to \infty}} \left(1 + \frac{1}{k}\right) e^{-1} = 1 \cdot e^{-1} = e^{-1}. \]

Step 7: Interpret the Limit

Since \( L = e^{-1} < 1 \), by the Ratio Test, the series is convergent.

Key Concepts

Convergence of Series

When analyzing the convergence of a series, we aim to determine whether the sum of all terms in the series reaches a finite number or continues indefinitely. In particular, an infinite series is a sum of the form \[ \sum_{k=1}^{\infty} a_k, \]where each \( a_k \) represents the terms in the series.### Types of Convergence- **Convergent Series:** If the sum approaches a specific value as the number of terms grows, the series converges.- **Divergent Series:** If the sum does not approach any value, or increases indefinitely, the series diverges.

• Determining convergence is important as many mathematical functions rely on understanding these sums.
• Various tests help in deciding convergence or divergence, such as the Ratio Test.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a^{x} \) where \( a \) is a constant base, and \( x \) is the exponent. These functions capture phenomena with rapid growth or decay.### Key Characteristics- **Growth and Decay:** Exponential functions can model growth, such as population increase, or decay, such as radioactive decay.- **Exponential Terms in Series:** In series, terms like \( e^{-k} \) indicate exponential decay.

• The base \( e \) is a special constant approximately equal to 2.71828, often used in natural logarithms and calculus.
• When used in a series, exponential terms frequently impact the convergence or divergence characteristics.

In our exercise, \( e^{-k} \) reduces rapidly as \( k \) increases, supporting the series' convergence.

Limit Computation

Calculating limits is crucial when evaluating the behavior of terms in a sequence or series as they approach infinity. The Ratio Test heavily relies on this concept, as it involves taking the limit of the ratio of consecutive terms.### Steps in Limit Computation1. **Substitution:** Incorporate larger and larger values to observe trends.2. **Simplification:** Simplify expressions, like recognizing that \( \frac{k+1}{k} \) simplifies to \( 1 + \frac{1}{k} \), facilitating limit calculation.3. **Limit Evaluation:** Calculate the final limit, which in Ratio Test informs about the series' nature.

• If the limit \( L < 1 \), it usually indicates convergence.
• If \( L > 1 \) or \( L = \infty \), this typically signals divergence.
• If \( L = 1 \), the test is inconclusive.

Limit computation gives insights into the behavior of terms as they stretch to infinity, vital in determining series' nature.