Problem 13 - Calculus: Early Transcendentals

Question

Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why. \( \displaystyle \sum_{n = 2}^{\infty} \frac {2}{n^2 - n} \)

Step-by-Step Solution

Answer

The series converges and the sum is 2.

Step 1: Identify the General Term

The sequence of terms is given by \( a_n = \frac{2}{n^2 - n} \). This can be simplified to \( a_n = \frac{2}{n(n-1)} \).

Step 2: Rewrite General Term for Partial Fractions

Rewrite the general term using partial fraction decomposition:\[ \frac{2}{n(n-1)} = \frac{A}{n} + \frac{B}{n-1} \]Equating the numerators:\[ 2 = A(n-1) + Bn \]Solving for \(A\) and \(B\), we get \(A = 2\) and \(B = -2\). Thus:\[ \frac{2}{n(n-1)} = \frac{2}{n} - \frac{2}{n-1} \]

Step 3: Compute Partial Sums

To find partial sums, consider the form derived in the previous step:\[ S_k = \sum_{n=2}^{k} \left( \frac{2}{n} - \frac{2}{n-1} \right) \]This series telescopes:\[ S_k = \left( \frac{2}{2} - \frac{2}{1} \right) + \left( \frac{2}{3} - \frac{2}{2} \right) + \ldots + \left( \frac{2}{k} - \frac{2}{k-1} \right) \]Most terms cancel, so:\[ S_k = 2\left(1 - \frac{1}{k}\right) \]

Step 4: Find Values of Partial Sums

Calculate the first 10 partial sums:When \(k=2\), \(S_2 = 0\); When \(k=3\), \(S_3 = 1\); When \(k=4\), \(S_4 = \frac{4}{3}\);Continue calculating up to \(S_{11}\).

Step 5: Graph Sequences

Plot both the sequence of terms \(a_n\) and the partial sums \(S_k\) on the same graph. The sequence of terms decreases but remains positive, while the partial sums show a pattern of convergence.

Step 6: Determine Convergence

Observe the partial sums as \(k\) increases. As \(k \to \infty\), \(S_k = 2\left(1 - \frac{1}{k}\right) \to 2\). Thus, the series converges to 2.

Key Concepts

Partial Sums

Partial sums are a way to simplify and analyze series. They consist of adding a finite number of terms from a given series. By observing how these sums evolve as more terms are added, we can glean insights into the behavior of the series as a whole.

In the exercise, we explored the series with the general term given by \( a_n = \frac{2}{n(n-1)} \). The task was to find at least 10 partial sums. To do this, we needed to recognize that each partial sum \( S_k \) was the cumulative addition of terms from the sequence starting with \( n=2 \) up to \( n=k \).

Given the formula for partial sums derived through partial fraction decomposition, the series telescopes, allowing many terms to cancel out. The partial sums for our series were found using \[ S_k = \sum_{n=2}^{k} \left( \frac{2}{n} - \frac{2}{n-1} \right) = 2\left(1 - \frac{1}{k}\right) \]. With this, calculating the first 10 partial sums was straightforward, providing clear values pointing to the series' overall behavior.

Sequence and Series

Sequences and series are fundamental concepts in calculus. A sequence is an ordered list of numbers generated by a specific rule. In this exercise, the sequence of terms was defined by \( a_n = \frac{2}{n(n-1)} \).

When these terms are added together, they create a series. Our focus in the exercise was on an infinite series that begins at \( n=2 \) and continues indefinitely. Understanding sequences helps in determining how the series behaves. Each term's size and pattern of growth, or decline, contribute to the behavior of the related series.

By writing the series as \( \sum_{n=2}^{\infty} \frac{2}{n(n-1)} \), we shifted our focus to analyzing an infinite collection of these terms. Through steps such as partial fraction decomposition, the complex series became easier to manage, which in this case led directly to a focus on evaluating partial sums.

Convergence and Divergence

Series in calculus are particularly interesting because of their convergence or divergence. A series converges if its partial sums approach a specific number as more terms are added; it diverges if they do not settle towards any finite value.

In our exercise, after calculating partial sums and analyzing the pattern, it became evident that the series converges. The formula \( S_k = 2\left(1 - \frac{1}{k}\right) \) for partial sums indicated that as \( k \to \infty \), \( S_k \to 2 \). This tells us the series converges to the sum of 2.

Understanding the distinction between convergence and divergence is essential. It determines if the series sums up to a finite value—useful in many mathematical and applied contexts—or if it grows without bound. In this exercise, the clear convergence towards a finite number emphasizes the importance of partial sums in reaching this conclusion.