Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{k=1}^{\infty} \frac{3^{k}+k}{k !}\)

Step-by-Step Solution

Answer

The sum is \( e^3 + e - 2 \).

Step 1: Break Down the Series

The given series is \( \sum_{k=1}^{\infty} \frac{3^{k}+k}{k !} \). This series is the sum of two distinct series: \( \sum_{k=1}^{\infty} \frac{3^{k}}{k !} \) and \( \sum_{k=1}^{\infty} \frac{k}{k !} \). We will evaluate each separately.

Step 2: Evaluate the Exponential Series

The first part, \( \sum_{k=1}^{\infty} \frac{3^{k}}{k !} \), can be recognized as part of the exponential series \( e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \). Hence, \( e^3 = \sum_{k=0}^{\infty} \frac{3^k}{k!} \). However, we start at \( k=1 \), so we subtract \( \frac{3^0}{0!} = 1 \), getting \( \sum_{k=1}^{\infty} \frac{3^k}{k!} = e^3 - 1 \).

Step 3: Evaluate the Harmonic Series

The second part, \( \sum_{k=1}^{\infty} \frac{k}{k !} \), simplifies using differentiation of \( omega\sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x \). Differentiate both sides with respect to \( x \) to get \( e^x = \sum_{k=1}^{\infty} \frac{k x^{k-1}}{k!} \). Setting \( x = 1 \), the series becomes \( e \sum_{k=1}^{\infty} \frac{1}{k!} = e \). So, \( \sum_{k=1}^{\infty} \frac{k}{k!} = e - 1 \).

Step 4: Combine the Results

Add the results from Steps 2 and 3. We found \( \sum_{k=1}^{\infty} \frac{3^k}{k!} = e^3 - 1 \) and \( \sum_{k=1}^{\infty} \frac{k}{k!} = e - 1 \). Therefore, the total sum is:\[(e^3 - 1) + (e - 1) = e^3 + e - 2\]

Step 5: Conclusion: Final Answer

The original series evaluates to \( e^3 + e - 2 \). We broke it down into two subsets, evaluated each using Taylor series knowledge, and combined them for the final result.

Key Concepts

Exponential Series

The exponential series is a powerful tool used to represent functions as infinite sums. It is most recognizable in the form of the exponential function, given by the series

• \[ e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} \]

In this series, each term involves powers of a number (here, \( x \)) divided by the factorial of the index. This formulation provides an approximation for the function \( e^x \) at any value of \( x \), and it's crucial for complex calculations in calculus and analysis.
The series converges for all values of \( x \), making it very versatile. If you notice, the structure of the exponential series makes it particularly useful for renewing its components in various mathematical contexts, as seen when distinguishing the part of the series involving \( 3^k \) from the original exercise. By recognizing the component series, one can substitute \( e^3 \) and adjust accordingly for different starting points within the sum.

Combining Series

Combining series involves taking apart a complicated series into two or more simpler components. The original series,

• \[ \sum_{k=1}^{\infty} \frac{3^{k}+k}{k!} \]

was split into two series. This method allows individual evaluation which can simplify the computation process considerably.
By looking at the original sum, we noticed it can be decomposed into

• \[ \sum_{k=1}^{\infty} \frac{3^k}{k!} \]
• \[ \sum_{k=1}^{\infty} \frac{k}{k!} \]

Once each part was individually evaluated, their results were then added together. This is useful for breaking complex series into manageable parts each solved using appropriate series knowledge, such as the exponential and a derivative series in this case.

Series Evaluation

Series evaluation is the process of calculating the sum of an infinite series. When faced with evaluating a series, the goal is to express the entire series in a form that has a known functional equivalent or is easily computable.
For example, the series

• \[ \sum_{k=1}^{\infty} \frac{3^k}{k!} \]

can be directly linked to the exponential function because of its format closely resembling the series for \( e^x \). Similarly, understanding series differentiation, as used in the evaluation of

• \[ \sum_{k=1}^{\infty} \frac{k}{k!} \]

can convert the problem into evaluating known simpler forms. By identifying a known series or simplifying the terms, evaluation becomes much simpler and more direct.

Taylor Series

The Taylor series is a mathematical tool for expressing functions as infinite polynomials centered around some point. This series provides a powerful method for approximating functions with potentially complex or unknown behaviors using simpler polynomial terms.
Each term of the polynomial involves derivatives of the function evaluated at a particular point. The exponential series itself is a prominent special case of Taylor series centered around zero (also known as Maclaurin series), which allows us to express exponential functions in simple infinite sums. A basic understanding of the Taylor series is instrumental in breaking down series and evaluating them by recognizing the structure relative to known expansions.