Problem 53 - Calculus for Biology and Medicine

Question

In Problems \(53-60\), write the first five terms of the sequence \(\left\\{a_{n}\right\\}\) \(n=0,1,2,3, \ldots\), and determine whether \(\lim _{n \rightarrow \infty} a_{n}\) exists. If the limit exists, find it. $$ a_{n}=\frac{n^{2}}{n+1} $$

Step-by-Step Solution

Answer

The sequence terms are 0, \(\frac{1}{2}\), \(\frac{4}{3}\), \(\frac{9}{4}\), and \(\frac{16}{5}\). The limit does not exist as \(n \to \infty\).

Step 1: Calculate the First Term

To find the first term, substitute \( n = 0 \) into the sequence formula \( a_n \). \[ a_0 = \frac{0^2}{0+1} = 0 \] So, the first term \( a_0 \) is 0.

Step 2: Calculate the Second Term

Next, substitute \( n = 1 \) into the sequence formula. \[ a_1 = \frac{1^2}{1+1} = \frac{1}{2} \] Thus, the second term \( a_1 \) is \( \frac{1}{2} \).

Step 3: Calculate the Third Term

Substitute \( n = 2 \) into the sequence formula. \[ a_2 = \frac{2^2}{2+1} = \frac{4}{3} \] Therefore, the third term \( a_2 \) is \( \frac{4}{3} \).

Step 4: Calculate the Fourth Term

Substitute \( n = 3 \) into the sequence formula. \[ a_3 = \frac{3^2}{3+1} = \frac{9}{4} \] Thus, the fourth term \( a_3 \) is \( \frac{9}{4} \).

Step 5: Calculate the Fifth Term

Substitute \( n = 4 \) into the sequence formula. \[ a_4 = \frac{4^2}{4+1} = \frac{16}{5} \] Therefore, the fifth term \( a_4 \) is \( \frac{16}{5} \).

Step 6: Determine the Limit as \( n \to \infty \)

To find the limit of \( a_n \) as \( n \to \infty \), consider the dominant terms in the numerator and denominator. The expression becomes: \[ \lim_{n \to \infty} \frac{n^2}{n+1} \] Divide both the numerator and the denominator by \( n \): \[ \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \] As \( n \to \infty \), \( \frac{1}{n} \to 0 \). Therefore, the limit is \( \lim_{n \to \infty} \frac{n}{1} = n \). Thus, the limit \( e \infty \) does not exist.

Key Concepts

Term Calculation

Understanding how to calculate the terms of a sequence is crucial when working with sequences and limits. In our example, we begin by evaluating the first few values of the sequence \( \left\{a_{n}\right\} \), defined by the formula \( a_n = \frac{n^2}{n+1} \). Let's see how it works for each of the first five terms:

• **First Term**: Insert \( n = 0 \) into the formula to get \( a_0 = \frac{0^2}{0+1} = 0 \). The first term is simply 0.


• **Second Term**: Insert \( n = 1 \) to find \( a_1 = \frac{1^2}{1+1} = \frac{1}{2} \). This gives us \( \frac{1}{2} \) for the second term.


• **Third Term**: Calculate \( a_2 \) by using \( n = 2 \). So, \( a_2 = \frac{2^2}{2+1} = \frac{4}{3} \), giving \( \frac{4}{3} \) as the third term.


• **Fourth Term**: For \( n = 3 \), \( a_3 = \frac{3^2}{3+1} = \frac{9}{4} \), which is \( \frac{9}{4} \).


• **Fifth Term**: Finally, use \( n = 4 \) to find \( a_4 = \frac{4^2}{4+1} = \frac{16}{5} \), resulting in \( \frac{16}{5} \).

Each term is calculated by plugging in a different \( n \) value into the given formula, showing how sequences serve to generalize patterns.

Infinite Sequence

Infinite sequences are essentially a list of numbers in a specific order that goes on forever. An infinite sequence is represented as \( \{a_0, a_1, a_2, \ldots \} \), where each term can be calculated using a formula. In the case of our sequence, \( a_n = \frac{n^2}{n+1} \), each term in the sequence is derived from a function of \( n \).

Working with infinite sequences involves understanding how the sequence behaves as it progresses towards infinity. Rather than merely looking at a small segment, infinite sequences help illustrate long-term behavior in mathematical modeling. By analyzing an infinite sequence, students can explore concepts that lay the groundwork for calculus and real analysis studies.

Limit Analysis

When analyzing limits of sequences, we seek to find the value that the elements of a sequence approach as \( n \) becomes very large. Here's how it works for our specific example of \( a_n = \frac{n^2}{n+1} \).

• We start with the formula for \( a_n \) and consider \( \lim_{n \to \infty} \frac{n^2}{n+1} \).


• By dividing both the numerator and the denominator by \( n \), simplify to \( \lim_{n \to \infty} \frac{n}{1 + \frac{1}{n}} \).


• As \( n \to \infty \), \( \frac{1}{n} \to 0 \), simplifying further to \( \lim_{n \to \infty} n \).

However, this does not produce a single finite value. It tells us the sequence grows without bound, meaning in this case, the limit does not exist as a finite number. In simpler terms, this sequence will keep increasing endlessly, without settling at any stable point, demonstrating how limits provide insight into infinite growth behavior.