Problem 33 - 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. $$a_{n}=\frac{24}{n^{3}}\left[\frac{n(n+1)(2 n+1)}{6}\right]$$

Step-by-Step Solution

Answer

The sequence is convergent and its limit is 8.

Step 1: Simplify the Expression Inside the Brackets

First, focus on the expression inside the brackets: \( \frac{n(n+1)(2n+1)}{6} \). This is the sum of the first \( n \) squares, \( \sum_{k=1}^{n} k^2 \), which is known to be equal to \( \frac{n(n+1)(2n+1)}{6} \).

Step 2: Substitute and Simplify

Substitute the expression back into the nth term of the sequence as follows:\[ a_n = \frac{24}{n^3} \times \frac{n(n+1)(2n+1)}{6} \].Simplifying this expression gives:\[ a_n = \frac{24n(n+1)(2n+1)}{6n^3} \].

Step 3: Simplify Further

Divide both the numerator and the denominator by 6:\[ a_n = \frac{4n(n+1)(2n+1)}{n^3} \].

Step 4: Expand and Simplify Further

Expand the numerator: \( n(n+1)(2n+1) = n(2n^2 + 3n + 1) = 2n^3 + 3n^2 + n \). Substituting back into the expression gives:\[ a_n = \frac{4(2n^3 + 3n^2 + n)}{n^3} = \frac{8n^3 + 12n^2 + 4n}{n^3} \].

Step 5: Divide Each Term by \( n^3 \)

Divide each term of the numerator by \( n^3 \):\[ a_n = 8 + \frac{12}{n} + \frac{4}{n^2} \].

Step 6: Determine the Limit

As \( n \to \infty \), \( \frac{12}{n} \to 0 \) and \( \frac{4}{n^2} \to 0 \). Thus:\[ \lim_{n \to \infty} a_n = 8 \]. The sequence is convergent, and its limit is 8.

Key Concepts

Convergent Sequences

A convergent sequence is a sequence where the terms approach a specific value as the sequence progresses towards infinity. As each term in the sequence gets larger, they zero in on this stable value called the limit. It means that after a certain point, the terms in the sequence will stay as close to the limit as you want them to, provided you move far enough along in the sequence. To understand better, think of a sequence like a chain of numbers. If it's convergent, then despite possibly starting off in different places, all numbers tend to head toward the same "destination." In the original exercise, the sequence reached a limit of 8.

Divergent Sequences

Divergent sequences do not settle down to a specific value. Instead, they either keep growing indefinitely, bounce around erratically, or perhaps slowly drift toward infinity. One of the most straightforward types of divergence is sequences going to infinity, like the sequence: 1, 2, 3, 4, and so on. This never approaches a fixed number but keeps increasing. In analyzing a sequence like in the original problem, a key step to determining whether it's divergent is to simplify the expression. If it doesn't reach a limit like 8 in previous instances, it may diverge. Remember, divergence is simply saying there isn’t just one value these numbers tend to gravitate toward. It could be endless growth, but it could also be endless ambiguity.

Limit of a Sequence

The limit of a sequence is the value that the terms of a sequence "approach" as the number of terms goes to infinity. By examining the limit of a sequence, we get insight into the sequence's behavior as it extends indefinitely.To find this limit, it often requires simplifying complicated expressions, as done in the exercise. You break them apart, without altering their value, to observe their core limiting behavior. Even seemingly complex sequences might simplify down nicely to a singular number.After simplification in the exercise, terms like \(\frac{12}{n}\) and \(\frac{4}{n^2}\) shrink to zero, clarifying that the sequence converges to a limit of 8. This clean observation confirms the pattern made apparent by the sequence’s tailing behavior.

Sum of Squares Formula

The sum of squares formula is a mathematical equation that simplifies the process of summing the squares of the first \(n\) natural numbers. This is given by the expression \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \). The formula makes it much simpler to evaluate the sum of squares efficiently, which would otherwise require adding up each squared term individually.In the original exercise, this formula was key to simplifying the given sequence term inside the brackets. By recognizing and utilizing this known formula, solving the exercise was greatly streamlined. This transformation significantly reduces calculation time and prevents mistakes by facilitating the recognition of complex patterns in sequences.