Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ 1-\frac{1}{2}, \frac{1}{2}-\frac{1}{3}, \frac{1}{3}-\frac{1}{4}, \frac{1}{4}-\frac{1}{5}, \ldots $$

Step-by-Step Solution

Answer

The explicit formula is \(a_n = \frac{1}{n} - \frac{1}{n+1}\) and the sequence converges to 0.

Step 1: Identify the Pattern

First, look for a pattern in the given sequence: \(1 - \frac{1}{2}, \frac{1}{2} - \frac{1}{3}, \frac{1}{3} - \frac{1}{4}, \ldots\). We observe that the general term for the sequence seems to be \(a_n = \frac{1}{n} - \frac{1}{n+1}\).

Step 2: Derive the Explicit Formula

Based on the pattern identified, the explicit formula is \(a_n = \frac{1}{n} - \frac{1}{n+1}\). Here, \(n\) is the term number starting from 1.

Step 3: Determine Convergence or Divergence

Check the limit as \(n\) approaches infinity. To determine if the sequence converges or diverges, calculate \(\lim_{n \to \infty} a_n\). In this case: \[ \lim_{n \to \infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) = \lim_{n \to \infty} \left( \frac{1}{n} \right) - \lim_{n \to \infty} \left( \frac{1}{n+1} \right)\] As \(n\) becomes very large, both terms approach 0. Thus, \(\lim_{n \to \infty} a_n = 0\). Therefore, the sequence converges.

Key Concepts

Explicit Formula of a Sequence

When given a sequence, finding the explicit formula is crucial as it allows us to express any term in the sequence directly with respect to its position. For the sequence given:

• first term: \(1 - \frac{1}{2}\)
• second term: \(\frac{1}{2} - \frac{1}{3}\)
• third term: \(\frac{1}{3} - \frac{1}{4}\)

we aim to identify a distinct pattern. By observing, we see that each term can be represented as \(a_n = \frac{1}{n} - \frac{1}{n+1}\). Here, \(n\) is simply the position of the term within the sequence, starting from 1.
This explicit formula captures the essence of how the terms change from one to the next, enabling us to calculate any term in the sequence without listing all preceding terms. This formula demonstrates the pattern of subtraction involving consecutive fractions, which is consistent across the sequence.

Limit of a Sequence

Understanding the limit of a sequence is crucial when determining whether a sequence converges or diverges as \(n\) approaches infinity. For the explicit formula \(a_n = \frac{1}{n} - \frac{1}{n+1}\), finding the limit involves evaluating the behavior of each fraction as \(n\) gets very large.
As \(n\) increases, both \(\frac{1}{n}\) and \(\frac{1}{n+1}\) approach zero. Therefore, we evaluate the limit of the whole expression:

• \(\lim_{n \to \infty} \frac{1}{n} = 0\)
• \(\lim_{n \to \infty} \frac{1}{n+1} = 0\)
• Thus, \(\lim_{n \to \infty} a_n = 0 - 0 = 0\)

This result indicates that the sequence converges to 0 as \(n\) becomes infinitely large. Thus, understanding the limit helps us conclude that the sequence does not veer off to infinity or settle elsewhere but instead steadies at zero.

Pattern Recognition in Sequences

Pattern recognition in sequences is often the initial step to solving problems involving sequences and series. Recognizing the pattern of sequence terms helps simplify finding formulas and understanding their behavior.
Looking at the sequence \(1 - \frac{1}{2}, \frac{1}{2} - \frac{1}{3}, \frac{1}{3} - \frac{1}{4}, \ldots\), the objective is to spot a repetitive structure. Here, we identify a pattern involving subtracting consecutive terms from the harmonic sequence \(\frac{1}{n}\).

• First, notice that each term subtracts a fraction that is one added to the denominator of its predecessor.
• This habit reflects in successive terms, creating a clear subtraction of terms of a form \(\frac{1}{n} - \frac{1}{n+1}\).

Recognizing these patterns is an essential skill, as it simplifies the complex appearance of sequences into understandable formulas. This skill is especially useful in advanced mathematics, where the essence of many problems lies in uncovering such patterns.