Problem 14 - 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. \( \displaystyle \sum_{n = 1}^{\infty} \left( \sin \frac {1}{n} - \sin \frac {1}{n + 1} \right) \)

Step-by-Step Solution

Answer

The series converges; the sum is \( \sin 1 \).

Step 1: Understand the Series

The series given is \( \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} - \sin \frac{1}{n+1} \right) \). This is a telescoping series because the terms involve a difference, which suggests cancellations will occur between successive terms.

Step 2: Identify the Partial Sums

To find the partial sums, calculate \( S_N = \sum_{n=1}^{N} \left( \sin \frac{1}{n} - \sin \frac{1}{n+1} \right) \). Observe that terms cancel out: \( \sin \frac{1}{1} - \sin \frac{1}{N+1} \).

Step 3: Calculate the First 10 Partial Sums

For the first 10 partial sums, calculate as follows:- \( S_1 = \sin \frac{1}{1} - \sin \frac{1}{2} \)- \( S_2 = \sin \frac{1}{1} - \sin \frac{1}{3} \)- Continue this process up to \( S_{10} = \sin \frac{1}{1} - \sin \frac{1}{11} \).

Step 4: Analyze the Partial Sums

The partial sum \( S_N \) simplifies to \( \sin 1 - \sin \frac{1}{N+1} \). As \( N \to \infty \), \( \sin \frac{1}{N+1} \to 0 \), so \( S_N \to \sin 1 \).

Step 5: Graph the Sequences

Graph the sequence of terms \( \sin \frac{1}{n} - \sin \frac{1}{n+1} \) and the sequence of partial sums \( S_N \). The graph of \( S_N \) should level off at \( \sin 1 \). The sequence of terms approaches zero as \( n \to \infty \), showing the series converges.

Step 6: Conclude on Convergence

Since the partial sums \( S_N \) approach \( \sin 1 \), the series is convergent. The sum of the series is \( \sin 1 \).

Key Concepts

Partial Sums

When working with series, the concept of partial sums is crucial. A partial sum is the sum of the first few terms of a series. It’s often denoted as \( S_N \), where \( N \) is the number of terms included. For the series provided, a telescoping series, the terms are structured so that many cancel each other out as you add them up.

In the exercise, the partial sum \( S_N \) of the series is derived by summing up to the \( N^{th} \) term:

• \( S_1 = \sin \frac{1}{1} - \sin \frac{1}{2} \)
• \( S_2 = \sin \frac{1}{1} - \sin \frac{1}{3} \)
• ... (continue up to \( S_{10} = \sin \frac{1}{1} - \sin \frac{1}{11} \))

As you can see, each new partial sum builds upon the previous one. This process highlights the beauty of telescoping series, as many pieces of the sequences cancel, simplifying the computation.

Series Convergence

Determining whether a series converges is fundamental in understanding its behavior in the long term. A series is said to converge if the sequence of its partial sums approaches a finite limit as more terms are added. In simpler words, it means that as we keep adding more terms, we approach a specific value.

For the series \( \sum_{n=1}^{\infty} \left( \sin \frac{1}{n} - \sin \frac{1}{n+1} \right) \), each term becomes smaller and smaller as \( n \) increases. As calculated, the partial sums \( S_N \) simplify to approximately \( \sin 1 - \sin \frac{1}{N+1} \). Here, \( \sin \frac{1}{N+1} \) approaches zero as \( N \) goes to infinity. Consequently, \( S_N \to \sin 1 \), showing that this series indeed converges, with the sum being \( \sin 1 \).

Sequence of Terms

The sequence of terms is like the building blocks of a series. Understanding the sequence helps in identifying the pattern and determining the convergence of the series.

In our specific series, the sequence of terms is given by \( \sin \frac{1}{n} - \sin \frac{1}{n+1} \). Each term represents the difference between consecutive reciprocal sines. As \( n \) becomes larger, the difference between \( \sin \frac{1}{n} \) and \( \sin \frac{1}{n+1} \) becomes smaller, heading towards zero. This behavior is typical in convergent telescoping series, where the terms shrink as the sequence progresses. Graphically representing this sequence alongside the partial sums allows us to visualize the cancellation effect, where each added term brings us a bit closer to the series's total sum. This pattern confirms that the series converges neatly to its sum, \( \sin 1 \).