Chapter 9: Problem 39
Find a formula for the \(n\) th partial sum of the series and use it to determine if the series converges or diverges. If a series converges, find its sum. $$\sum_{n=1}^{\infty}\left(\cos ^{-1}\left(\frac{1}{n+1}\right)-\cos ^{-1}\left(\frac{1}{n+2}\right)\right)$$
Short Answer
Expert verified
The series converges with a sum of \(-\frac{\pi}{6}\).
Step by step solution
01
Identify the Series Type
Observe that the series in question is of the form \( \sum_{n=1}^{\infty} \left(a_n - a_{n+1}\right) \), where \( a_n = \cos^{-1}\left(\frac{1}{n+1}\right) \). This is a telescoping series.
02
Determine the n-th Term of the Partial Sum
The \( n \)th partial sum \( S_n \) of a telescoping series is found by adding the first \( n \) terms: \( S_n = \sum_{k=1}^{n} \left( a_k - a_{k+1} \right) \). For this series, it simplifies to \( a_1 - a_{n+1} \).
03
Calculate the First Term of the Series
Find \( a_1 \), which is \( \cos^{-1}\left(\frac{1}{2}\right) \). This is equal to \( \frac{\pi}{3} \) since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).
04
Express the n-th Partial Sum
Substitute the value of \( a_1 \) into the expression for the partial sum: \( S_n = \frac{\pi}{3} - \cos^{-1}\left(\frac{1}{n+2}\right) \).
05
Determine the Limit of the Partial Sum as n Approaches Infinity
Calculate \( \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{\pi}{3} - \cos^{-1}\left(\frac{1}{n+2}\right) \right) = \frac{\pi}{3} - \cos^{-1}(0) = \frac{\pi}{3} - \frac{\pi}{2} = -\frac{\pi}{6} \).
06
Analyze Convergence
Since the limit \( \lim_{n \to \infty} S_n \) exists and is finite, the series converges. The sum of the series is \(-\frac{\pi}{6}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Sum
A partial sum in a series is the sum of the first few terms, up to the th term. It provides a way to estimate the total sum of an infinite series by looking at a finite part of it. Understanding partial sums helps in analyzing whether an infinite series converges or diverges.
For a telescoping series like in our exercise, the partial sum has a specific pattern. Each term cancels out a neighboring term, simplifying the calculation. In our example, the partial sum for the first terms is given by:
For a telescoping series like in our exercise, the partial sum has a specific pattern. Each term cancels out a neighboring term, simplifying the calculation. In our example, the partial sum for the first terms is given by:
- Calculate the sum up to the th term: \[ S_n = a_1 - a_{n+1} \] where \( a_n = \cos^{-1}\left( \frac{1}{n+1} \right) \).
Convergence and Divergence
In the context of infinite series, understanding convergence and divergence is essential. A series converges if the partial sums approach a finite limit as you add more terms.
Conversely, a series diverges if there is no finite limit. Some series might increase indefinitely or oscillate without settling.
Conversely, a series diverges if there is no finite limit. Some series might increase indefinitely or oscillate without settling.
- Convergence: When the limit of partial sums is finite, the series converges. It means the sum tends to a specific value. In our example, the limit \( \lim_{n \to \infty} S_n = -\frac{\pi}{6} \) shows convergence.
- Divergence: If no finite limit exists, the series diverges. The sums could go to infinity or have no pattern to settle.
Arccosine Function
The arccosine function, denoted as \( \cos^{-1}(x) \), is the inverse of the cosine function. It returns the angle whose cosine value is \( x \). Understanding this function is crucial in our exercise since it dictates the series terms. The range of arccosine is from \( 0 \) to \( \pi \), ensuring the output is an angle.
Let’s look at how it worked in the problem:
As you use the arccosine function, think about its inverse nature and range, giving you a geometrical perspective on angle measurements.
Let’s look at how it worked in the problem:
- The arccosine values, such as \( \cos^{-1}\left( \frac{1}{n+1} \right) \), form the building blocks of the telescoping series.
- We find specific values like \( a_1 = \cos^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{3} \), providing a starting point.
As you use the arccosine function, think about its inverse nature and range, giving you a geometrical perspective on angle measurements.