Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the series given in Problems \(27-32\), determine how large \(n\) must be so that using the nth partial sum to approximate the series gives an error of no more than \(0.0002 .\) \(\sum_{k=1}^{\infty} \frac{1}{k(k+1)}\)

Step-by-Step Solution

Answer

The partial sum for \(n = 5000\) gives an error of less than 0.0002.

Step 1: Recognize Series Type

The series \(\sum_{k=1}^{\infty} \frac{1}{k(k+1)}\) is a telescoping series. In a telescoping series, each term canceled part of earlier terms, leading to many terms vanishing upon summation.

Step 2: Decompose Fraction

Decompose the term \(\frac{1}{k(k+1)}\) using partial fraction decomposition: \(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\). This sets up for cancellation when summed.

Step 3: Write the Partial Sum Formula

Write the \(n\)-th partial sum formula for the series based on partial fraction decomposition. The partial sum \(S_n = 1 - \frac{1}{n+1}\) follows from how terms cancel in the series expansion.

Step 4: Determine Final Answer for n

To ensure the absolute error is no more than 0.0002, solve \(\left|S - S_n\right| < 0.0002\). Since \(S_n = 1 - \frac{1}{n+1}\) and \(S = 1\), this becomes \(\frac{1}{n+1} < 0.0002\). Hence, solve \(n+1 > 5000\), leading to \(n > 4999\).

Step 5: Verify Calculation

Since \(n > 4999\), rounding to the nearest whole, \(n = 5000\) satisfies \(\frac{1}{5001} \approx 0.00019996 < 0.0002\). Therefore, with \(n = 5000\), the condition \(\left|S - S_n\right| < 0.0002\) is met.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in algebra to break down complex rational expressions into simpler fractions. This method is especially helpful when dealing with series as it sets the stage for terms to cancel out, simplifying the overall expression.
For instance, in the series \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \), we notice that each term is not easily summable in its original form. To tackle this, we perform partial fraction decomposition:

• Start with \( \frac{1}{k(k+1)} \).
• Express it as \( \frac{1}{k} - \frac{1}{k+1} \).

This transformation is incredibly useful as it makes each term simple to handle and reveals a telescoping nature, facilitating the summation process.

Partial Sum

The partial sum of a series refers to the sum of the first \( n \) terms. It provides an approximation to the total sum of an infinite series. Calculating the partial sum is essential for assessing series convergence and for practical computations.
In the case of our telescoping series, after applying partial fraction decomposition, we easily compute the partial sum \( S_n \). For this series, the partial sum after the decomposition is:

• \( S_n = 1 - \frac{1}{n+1} \)

This result comes from the telescoping nature, where most terms cancel out, leaving us only with the first term and the negative of the term at \( k = n+1 \). As \( n \) grows, \( \frac{1}{n+1} \) becomes negligible, approximating the actual sum.

Series Convergence

Series convergence indicates whether the sum of an infinite series approaches a finite limit as more terms are added. It's an important aspect to understand before attempting to approximate or compute series sums.
For a telescoping series like \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \), convergence can be seen clearly through the cancellation in partial sums. As \( n \) tends towards infinity, \( S_n = 1 - \frac{1}{n+1} \) essentially becomes 1, since

• \( \frac{1}{n+1} \to 0 \) as \( n \to \infty \)

This direct cancellation and approach to a finite sum (1) shows that the series is convergent, giving us confidence to approximate it accurately.

Error Estimation

Error estimation is crucial when using a finite number of terms to approximate the sum of an infinite series. It helps ensure the approximation is within a certain acceptable range, known as the error bound.
For the given series, we want our error to be no greater than 0.0002. The error bound can be articulated in terms of the difference between the infinite sum and the nth partial sum:

• \( \left| S - S_n \right| = \frac{1}{n+1} \)

To keep this error less than 0.0002, solve the inequality:

• \( \frac{1}{n+1} < 0.0002 \)
• Thus, \( n+1 > 5000 \) which simplifies to \( n > 4999 \)

Choosing an \( n \) of 5000 ensures that the partial sum is a good approximation, and the actual error is small enough to meet the requirement, confirming that our approach is reliable.