Problem 29 - Precalculus: Mathematical for Calculus

Question

If the sequence is convergent, find its limit. If it is divergent, explain why. $$a_{n}=\frac{24}{n^{3}}\left[\frac{n(n+1)(2 n+1)}{6}\right]$$

Step-by-Step Solution

Answer

The sequence is convergent, and its limit is 8.

Step 1: Identify the Sequence

The given sequence is defined as \( a_n = \frac{24}{n^3} \left[ \frac{n(n+1)(2n+1)}{6} \right] \). First, notice the expression inside the brackets \( \frac{n(n+1)(2n+1)}{6} \). This represents the formula for the sum of the squares of the first \( n \) natural numbers, which is \( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \).

Step 2: Simplify the Sequence

Substitute the known formula for the sum of squares into the sequence definition. Thus, the sequence becomes \( a_n = \frac{24}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \). Simplify this expression by multiplying the constants: \( a_n = \frac{4n(n+1)(2n+1)}{n^3} \).

Step 3: Expand and Simplify Further

Expand the expression in the numerator: \( n(n+1)(2n+1) = 2n^3 + 3n^2 + n \). Substitute this back into the sequence: \( a_n = \frac{4(2n^3 + 3n^2 + n)}{n^3} \).

Step 4: Divide Each Term by \(n^3\)

Split the terms in the fraction individually: \( a_n = \frac{4 \cdot 2n^3}{n^3} + \frac{4 \cdot 3n^2}{n^3} + \frac{4n}{n^3} \). Simplifying each term gives: \( a_n = 8 + \frac{12}{n} + \frac{4}{n^2} \).

Step 5: Analyze the Limit as \(n\) Approaches Infinity

As \( n \to \infty \), the terms \( \frac{12}{n} \) and \( \frac{4}{n^2} \) both approach zero. Therefore, the limit of \( a_n \) as \( n \to \infty \) is \( 8 + 0 + 0 = 8 \).

Step 6: Conclusion About Convergence

The sequence \( a_n \) approaches 8 as \( n \) goes to infinity. Therefore, the sequence is convergent and its limit is 8.

Key Concepts

Sequence Convergence

When we talk about sequence convergence, we are discussing whether the sequence approaches a specific value as it progresses towards infinity. In mathematical terms, if a sequence \( \{a_n\} \) converges, it means there exists some fixed value \( L \) such that as \( n \) becomes very large, \( a_n \) gets arbitrarily close to \( L \). This value \( L \) is known as the limit of the sequence.

The sequence in question, which is \( a_n = \frac{4n(n+1)(2n+1)}{n^3} \), has been simplified to \( a_n = 8 + \frac{12}{n} + \frac{4}{n^2} \). As \( n \) approaches infinity, the terms \( \frac{12}{n} \) and \( \frac{4}{n^2} \) vanish. This is because dividing by a large number makes the fraction's value approach zero.

• The term \( 8 \) remains unaffected by \( n \), and thus the sequence converges to \( 8 \).
• Hence, the sequence is said to be convergent with a limit of \( 8 \).

Sum of Squares

The sum of squares is a fundamental concept in mathematics. It represents the addition of the squares of the first \( n \) natural numbers. The formula for this, \( \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} \), is pivotal as it transforms a potentially complex calculation into a simple expression.

In the original exercise, the sum of squares formula \( \frac{n(n+1)(2n+1)}{6} \) is used as part of the sequence \( a_n \). Essentially, this is embedded in the sequence formula, simplifying its complexity. By substituting the sum of squares into our sequence definition, we can reframe the sequence in a manner that is manageable and easy to analyze.

• This substitution is vital, as it allows us to simplify the sequence and reveal the convergent behavior by showing how higher power terms dominate the numerator.

Asymptotic Behavior

Asymptotic behavior describes how a sequence behaves as it approaches infinity, essentially providing insights into the trend for very large values of \( n \). In our sequence \( a_n = 8 + \frac{12}{n} + \frac{4}{n^2} \), the terms \( \frac{12}{n} \) and \( \frac{4}{n^2} \) demonstrate crucial aspects of asymptotic behavior.

As \( n \) increases:

• The term \( \frac{12}{n} \) shrinks towards zero proportionally to \( n^{-1} \).
• Similarly, \( \frac{4}{n^2} \) becomes negligible even faster, decreasing like \( n^{-2} \).

The dominance of the constant term "8" becomes evident because the diminishing fractions become insignificant. Hence, asymptotic analysis reveals that while the whole sequence appears complex, its behavior is overwhelmingly governed by the constant component. This understanding helps to anticipate the outcome of \( a_n \) as \( n \to \infty \), reinforcing the concept of sequence convergence.