Problem 38 - Calculus: Early Transcendentals

Question

Find the sum of the series \( \sum_{n = 1}^{\infty} ne^{-2n} \) correct to four decimal places.

Step-by-Step Solution

Answer

The sum of the series is approximately 0.1808.

Step 1: Identify the Type of Series

The given series is of the form \( \sum_{n=1}^{\infty} ne^{-2n} \). This indicates it's an infinite series involving an exponential function.

Step 2: Use Known Series Techniques

Recognize that \( e^{-2n} \) is an exponential decay function. Use the known series sum for \( \sum_{n=1}^{\infty} nx^n \), which is \( \frac{x}{(1-x)^2} \) for \( |x| < 1 \). Substitute \( x = e^{-2} \) into this formula.

Step 3: Substitute and Simplify

Substitute \( x = e^{-2} \) into the series sum formula:\[\sum_{n=1}^{\infty} ne^{-2n} = \frac{e^{-2}}{(1-e^{-2})^2}\]

Step 4: Calculate the Values Numerically

Evaluate \( e^{-2} \) approximately, which is about 0.1353. Insert this into the formula:\[\frac{0.1353}{(1 - 0.1353)^2} = \frac{0.1353}{0.865\times0.865} \] Calculate further to find the sum.

Step 5: Compute the Final Value

Calculate the denominator: \( 0.865 \times 0.865 \approx 0.748225 \) and substitute:\[\frac{0.1353}{0.748225} \approx 0.1808\] Thus, the sum of the series is approximately 0.1808.

Key Concepts

Exponential Function

An exponential function is a type of mathematical function where a constant base is raised to a variable exponent, like in the expression \( e^{-2n} \). The letter \( e \) represents Euler's number, which is an irrational number approximately equal to 2.71828. Exponential functions are key in modelling real-world phenomena like growth and decay. When the exponent is negative, as in \( e^{-2n} \), it signifies a decay function, meaning the values decrease as \( n \) increases. This behavior is crucial for understanding how the values in the series behave as it sums over infinite terms.

Series Sum Formula

To find the sum of an infinite series like \( \sum_{n=1}^{\infty} ne^{-2n} \), we make use of known series sum formulas. In this case, the relevant formula is \( \sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} \), which applies for values of \( |x| < 1 \). This formula is extremely useful as it gives us a way to summate infinite terms efficiently and accurately.

When \( x \) is replaced with \( e^{-2} \), we transform our series into something more manageable:

• Substitute \( x = e^{-2} \) into the formula:

\[ \sum_{n=1}^{\infty} ne^{-2n} = \frac{e^{-2}}{(1-e^{-2})^2} \] This transforms the infinite series into a much simpler expression that can be evaluated with straightforward calculations.

Numerical Evaluation

Numerical evaluation involves calculating approximate values for complex mathematical expressions. For our series, the expression reduces to:

• \( e^{-2} \approx 0.1353 \)
• Substitute into the series sum formula:

\[ \frac{0.1353}{(1 - 0.1353)^2} \]Performing the steps, we calculate:

• The denominator \( 0.865 \times 0.865 \approx 0.748225 \)
• Finally, evaluate the entire expression:

\[ \frac{0.1353}{0.748225} \approx 0.1808 \] This numerical approximation gives us the value of the series sum correct to four decimal places. It demonstrates the power of numerical techniques in solving complex mathematical problems in a concise and precise manner.