Problem 25 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = \frac {n^4}{n^3 - 2n} \)

Step-by-Step Solution

Answer

The sequence diverges.

Step 1: Simplify the Fraction

The sequence is given by \( a_n = \frac {n^4}{n^3 - 2n} \). To simplify this, divide the numerator and the denominator by \( n^3 \), the highest power of \( n \) in the denominator. Therefore, \[a_n = \frac{n^4}{n^3 - 2n} = \frac{\frac{n^4}{n^3}}{\frac{n^3}{n^3} - \frac{2n}{n^3}} = \frac{n}{1 - \frac{2}{n^2}}.\]

Step 2: Examine the Limit as n Approaches Infinity

As \( n \to \infty \), \( \frac{2}{n^2} \to 0 \), so the expression \( 1 - \frac{2}{n^2} \to 1 \). Rewrite the expression as: \[a_n = \frac{n}{1 - \frac{2}{n^2}} o \frac{n}{1} = n.\]

Step 3: Determine Convergence or Divergence

Since \( a_n = n \) as \( n \to \infty \), the expression tends to infinity. Therefore, the sequence diverges because it does not approach a finite limit.

Key Concepts

Limits in Calculus

In calculus, limits are fundamental in understanding how sequences and functions behave as they approach certain points. When we talk about the limit of a sequence, we mean the value that the terms of the sequence get closer to as the index of the sequence (usually denoted by \( n \)) becomes very large. For a sequence to converge, the terms must approach a specific, finite number.
Often, calculating the limit involves simplifying the algebra in an expression and observing how the sequence behaves as \( n \) approaches infinity. This is essential for understanding convergence. In some cases, the sequence values approach infinity rather than a particular number. Such scenarios indicate divergence because there is no finite limit.
When dealing with polynomials within sequences, finding the limit can require dividing both the numerator and denominator by the highest power of \( n \) in the denominator or using other algebraic manipulations to simplify the expression. This step makes it easier to analyze how the sequence behaves as \( n \) approaches infinity.

Infinite Sequences

An infinite sequence extends indefinitely, meaning there is no largest term. Infinite sequences can exhibit different behaviors based on their terms' tendencies as they advance.

• Convergent Sequences: Where the terms approach a specific finite number, making them converge to a particular limit. For instance, a simple geometric sequence with a rapidly decreasing ratio can converge to zero.
• Divergent Sequences: In these cases, the terms do not approach any finite number, often growing infinitely large. For example, sequences like \( a_n = n \) diverge because their values increase towards infinity.

Understanding infinite sequences demands examining the behavior of terms as \( n \) grows. If the terms stabilize around a finite number, the sequence converges. If not, we have divergence, as found in the original exercise where \( a_n = n \) diverges by growing boundlessly.

Dividing Polynomials in Sequences

When examining sequences that include polynomials, dividing them can help determine convergence or divergence. Polynomials are expressions consisting of variables and coefficients, involving operations like addition, subtraction, multiplication, and non-negative integer exponents of variables.
To analyze the sequence \( a_n = \frac{n^4}{n^3 - 2n} \), the process began by factoring out the largest power of \( n \) in the denominator, which is \( n^3 \). This resulted in a simpler expression \( \frac{n}{1 - \frac{2}{n^2}} \).
Such simplification reveals how the sequence behaves as \( n \) approaches infinity. In this instance, as \( n \) grows, \( \frac{2}{n^2} \) tends to zero, and we examine the expression \( \frac{n}{1} = n \). It's clear that as \( n \) increases, so do the term values.* Therefore, the sequence diverges as there is no finite limit, akin to our previous classification of divergent infinite sequences.