Problem 31 - 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. $$a_{n}=\frac{3}{n^{2}}\left[\frac{n(n+1)}{2}\right]$$

Step-by-Step Solution

Answer

The sequence is convergent, and its limit is \(\frac{3}{2}\).

Step 1: Simplify the Expression Inside the Brackets

Given the sequence term \(a_n = \frac{3}{n^2}\left[\frac{n(n+1)}{2}\right]\), start by simplifying the expression inside the brackets. Recognize that \(\frac{n(n+1)}{2}\) is the formula for the \(n\)-th triangular number.

Step 2: Simplify the Triangular Number Formula

The \(n\)-th triangular number can be rewritten as \(T_n = \frac{n^2 + n}{2}\). Substitute this back into the sequence term: \(a_n = \frac{3}{n^2} \times \frac{n^2 + n}{2}\).

Step 3: Multiply and Simplify

Multiply the terms: \(a_n = \frac{3(n^2 + n)}{2n^2} = \frac{3n^2 + 3n}{2n^2}\). Simplify this to get \(a_n = \frac{3n^2}{2n^2} + \frac{3n}{2n^2} = \frac{3}{2} + \frac{3}{2n}\).

Step 4: Determine the Limit

As \(n\) approaches infinity, the term \(\frac{3}{2n}\) approaches 0. Therefore, \(a_n = \frac{3}{2} + \frac{3}{2n}\) approaches \(\frac{3}{2}\).

Step 5: State the Convergence

Since the remaining term approaches a specific constant (\(\frac{3}{2}\)), the sequence \(a_n\) is convergent.

Key Concepts

Convergent Sequences

In mathematics, a convergent sequence is one where the terms approach a specific value as they progress to infinity. Think of it like a car steadily driving towards a destination. As you keep going further, your position gets closer and closer to the target location. In simpler terms, a sequence \(a_n\) is convergent if there exists a number \L\ such that for any given small distance, we can find a point in the sequence beyond which all the terms are within this small distance from \L\. This magical number \L\ is called the limit of the sequence. In the given exercise, the sequence \( a_n = \frac{3}{2} + \frac{3}{2n} \) converges to \(\frac{3}{2}\). As \(n\) becomes very large, the term \(\frac{3}{2n}\) becomes negligible, effectively making the distance between \(a_n\) and \(\frac{3}{2}\) disappear. Thus, the sequence closes in on \(\frac{3}{2}\), illustrating the charming concept of convergence.

Triangular Numbers

Triangular numbers are a sequence of numbers that can form an equilateral triangle. These numbers are named for their triangular shape when arranged in dots. The mathematical expression for the \(n\)-th triangular number is \([T_n = \frac{n(n+1)}{2}]\).

• The first triangular number \(T_1\) is 1.
• The second triangular number \(T_2\) is 3. Arrange 1 at the top and then a row of 2 below to form a small triangular shape.
• The third triangular number \(T_3\) is 6, adding another row of 3 dots.

The exercise we looked at involved triangular numbers because the given sequence formula included the term \( \frac{n(n+1)}{2} \). Recognizing triangular numbers is a crucial skill, as it allows for simplifications like we saw in the original step-by-step solution. Once recognized, it allows further exploration of relationships within the sequence.

Infinite Limit

An infinite limit describes the behaviour of a sequence as it progresses indefinitely. This is different from the sequence reaching a specific constant value, commonly termed its limit. Infinite limits address cases where sequences either climb endlessly towards infinity or drop without bound towards negative infinity. While our exercise resulted in a convergent sequence, understanding infinite limits is equally important, especially when sequences don't settle at a single value. A sequence with an infinite limit doesn't stabilize, it just keeps increasing or decreasing without ever settling. In the solution, the sequence \(a_n\) did not have an infinite limit, because \(a_n\) approached the finite value \(\frac{3}{2}\). However, recognizing that sequences can also tend to infinity helps expand our understanding of sequence behavior overall.