Problem 23 - Precalculus: Mathematics for Calculus

Question

Limits of Sequences If the sequence with the given \(n\) th term is convergent, find its limit. If it is divergent, explain why. $$a_{n}=\frac{1+n}{n+n^{2}}$$

Step-by-Step Solution

Answer

The sequence is convergent with a limit of 0.

Step 1: Identify the nature of the sequence

The given sequence is \(a_n = \frac{1+n}{n+n^2}\). Our goal is to find the limit as \(n\) approaches infinity. To do this, we must analyze the behavior of the sequence as \(n\) becomes very large.

Step 2: Simplify the expression

Let's simplify \(a_n = \frac{1+n}{n+n^2}\). Divide both the numerator and the denominator by \(n^2\), the highest power of \(n\) in the denominator:\[ a_n = \frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n} + 1} \]

Step 3: Take the limit as \(n\) approaches infinity

Evaluate the limit:\[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n} + 1} \]As \(n\) approaches infinity, \(\frac{1}{n^2}\) and \(\frac{1}{n}\) both approach 0. Thus, the expression simplifies to:\[ \frac{0+0}{0+1} = 0 \]

Step 4: Conclusion about convergence

Since the limit of the sequence exists and is a finite number, \(a_n = \frac{1+n}{n+n^2}\) is convergent. The limit is \(0\).

Key Concepts

Sequence Convergence

When discussing sequence convergence, we are essentially exploring whether a sequence approaches a specific value as the number of terms increases indefinitely. A sequence is said to converge if, as we progress through its terms to infinity, it becomes arbitrarily close to a certain number, known as the limit. Sequences that do not settle at any particular value are termed divergent.

In the given exercise, our sequence is defined as \(a_n = \frac{1+n}{n+n^2}\). The task is to determine if this sequence converges and, if so, to find the limit. By examining the behavior of the sequence as \(n\) increases, we can ascertain its convergence and conclude whether it is heading towards a specific limit.

Evaluating Limits

Evaluating limits is a crucial skill in understanding the long-term behavior of sequences. It involves finding the value that a sequence approaches as its index tends toward infinity. There are several techniques to evaluate limits, but they often require transforming the sequence into a form where the limit is more apparent.

In this exercise, we simplify the sequence \(a_n = \frac{1+n}{n+n^2}\) by dividing the numerator and the denominator by \(n^2\), the highest power present in the denominator. This way, we aim to eliminate terms that become insignificant as \(n\) tends to infinity. This simplification is an effective strategy to evaluate limits of sequences involving rational expressions.

Dividing by Highest Power

This strategy is often used when dealing with sequences and functions to simplify the evaluation of limits. By dividing the entire expression by the highest power of \(n\) present in the denominator, the expression becomes easier to analyze as \(n\) becomes very large.

For our sequence, \(a_n = \frac{1+n}{n+n^2}\), dividing by \(n^2\) gives \(a_n = \frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n} + 1}\). This transforms the original sequence into a format where as \(n\) approaches infinity, terms like \(\frac{1}{n}\) and \(\frac{1}{n^2}\) move towards zero, significantly simplifying the limit evaluation process.

Infinite Limits

Infinite limits occur when a sequence or function grows without bound as it approaches a particular point. However, in our context where \(n\) tends toward infinity, it is vital to determine whether a sequence approaches a finite limit or diverges.

In the given exercise, as we take the limit of \(a_n = \frac{1+n}{n+n^2}\), we observe that the transformed sequence \(\frac{\frac{1}{n^2} + \frac{1}{n}}{\frac{1}{n} + 1}\) simplifies to \(\frac{0 + 0}{0 + 1}\), which equals 0 as \(n\) approaches infinity. This outcome indicates that our sequence does not exhibit an infinite limit, but instead converges to a finite value of 0.