Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

An explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{\cos (n \pi)}{n} $$

Step-by-Step Solution

Answer

The sequence is \(-1, \frac{1}{2}, \frac{-1}{3}, \frac{1}{4}, \frac{-1}{5}\), it converges, and the limit is 0.

Step 1: Calculate the First Five Terms

To find the first five terms of the sequence defined by \(a_n = \frac{\cos(n \pi)}{n}\), substitute \(n = 1, 2, 3, 4, 5\) into the formula. - When \(n = 1\), \(a_1 = \frac{\cos(1 \pi)}{1} = \frac{-1}{1} = -1\). - When \(n = 2\), \(a_2 = \frac{\cos(2 \pi)}{2} = \frac{1}{2}\). - When \(n = 3\), \(a_3 = \frac{\cos(3 \pi)}{3} = \frac{-1}{3}\). - When \(n = 4\), \(a_4 = \frac{\cos(4 \pi)}{4} = \frac{1}{4}\). - When \(n = 5\), \(a_5 = \frac{\cos(5 \pi)}{5} = \frac{-1}{5}\). The first five terms are: \(-1, \frac{1}{2}, \frac{-1}{3}, \frac{1}{4}, \frac{-1}{5}\).

Step 2: Determine Convergence or Divergence

The sequence is given by \(a_n = \frac{\cos(n \pi)}{n}\), where the numerator, \(\cos(n \pi)\), alternates between \(-1\) and \(1\). As \(n\) increases, the denominator \(n\) grows larger. Hence, each term \(|a_n| = \frac{1}{n}\) becomes smaller as \(n\) increases. Since \(\frac{1}{n}\) converges to \(0\) as \(n\) goes to infinity, \(a_n\) also converges to \(0\).

Step 3: Find the Limit of the Sequence

Given that \(a_n = \frac{\cos(n \pi)}{n}\) converges, calculate \(\lim_{n \to \infty} a_n\). Since the magnitude \(\left|a_n\right| = \frac{1}{n}\) approaches \(0\) as \(n\) grows larger, the sequence converges to \(0\). Therefore, \(\lim_{n \to \infty} a_n = 0\).

Key Concepts

Limit of a Sequence

In calculus, the limit of a sequence defines the value that the terms of the sequence approach as the index (usually denoted by \(n\)) goes towards infinity. Finding this limit helps determine the behavior of the sequence over a long run.
For the sequence \(a_n = \frac{\cos(n \pi)}{n}\), our task is to determine if the sequence converges into a specific number or diverges by not approaching any single value. Here, convergence means the terms of the sequence get closer to a specific number, in this case, zero, as \(n\) grows larger.
The numerator alternates between -1 and 1 due to \(\cos(n \pi)\), while the denominator \(n\) increases indefinitely. Since \(|a_n| = \frac{1}{n}\), we're analyzing \(\frac{1}{n}\), a familiar form known to converge to 0. Thus, \(a_n\) also converges to 0.

Alternating Sequence

An alternating sequence is one where the terms switch between positive and negative. This back-and-forth nature can often aid in canceling out parts of the sequence when looking for convergence.
In our sequence, we have \(a_n = \frac{\cos(n \pi)}{n}\). The cosine component, \(\cos(n \pi)\), generates this alternating behavior where:

• \(\cos(n \pi) = -1\) for odd \(n\)
• \(\cos(n \pi) = 1\) for even \(n\)

Due to this phenomenon, the terms of our sequence become negative and positive alternately, such as -1, \(\frac{1}{2}\), \(-\frac{1}{3}\), \(\frac{1}{4}\)... This alternating nature combined with the denominator \(n\) growing larger ensures the sequence settles to zero over time.

Calculus Sequence Analysis

Analyzing a sequence in calculus involves understanding how the terms behave as you proceed towards infinity. We break down the given function and apply calculus principles to determine the limit or divergence of the sequence.

Our sequence is \(a_n = \frac{\cos(n \pi)}{n}\). To analyze, we first check the individual components of the sequence:

• Numerator: \(\cos(n \pi)\) produces an alternating sign pattern but remains bounded.
• Denominator: \(n\) one increasing without bound, leading to values diminishing in magnitude.

We then consider \(|a_n| = \frac{1}{n}\), which descends towards thin nothingness as \(n\) becomes very large. By understanding these behaviors through sequence analysis, we confirm \(\lim_{n \to \infty} a_n = 0\). This type of intuitive analysis is a cornerstone in exploring sequences in calculus.