Problem 31 - Precalculus Mathematics for Calculus

Question

If the sequence 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 converges, and its limit is \( \frac{3}{2} \).

Step 1: Simplify the Sequence Expression

We start by simplifying the given sequence expression: \(a_n = \frac{3}{n^2} \left[ \frac{n(n+1)}{2} \right] \). The inner expression \( \frac{n(n+1)}{2} \) represents the nth triangular number, which simplifies to: \( \frac{1}{2} (n^2 + n) \). Thus, the sequence becomes: \[ a_n = \frac{3}{n^2} \times \frac{n^2 + n}{2} = \frac{3}{2n^2} (n^2 + n). \]

Step 2: Distribute and Combine Terms

Distribute the 3 over the terms in the expression: \( \frac{3}{2n^2}(n^2 + n) = \frac{3n^2 + 3n}{2n^2} \). Thus, the sequence simplifies further to: \[ a_n = \frac{3n^2}{2n^2} + \frac{3n}{2n^2} = \frac{3}{2} + \frac{3}{2n}. \]

Step 3: Analyze the Limit of the Sequence

Now, consider the sequence \(a_n = \frac{3}{2} + \frac{3}{2n}\). As \(n\) approaches infinity, \(\frac{3}{2n}\) approaches zero because the denominator grows larger. Therefore, the expression simplifies to \( a_n \rightarrow \frac{3}{2} + 0 = \frac{3}{2} \).

Step 4: Conclude Convergence and Find Limit

Having determined that the \( \frac{3}{2n} \) term vanishes as \(n\to\infty\), we conclude the sequence \(a_n\) converges. The limit is \( \frac{3}{2} \).

Key Concepts

Limit of a Sequence

In calculus, the concept of the limit of a sequence is fundamental for understanding how sequences behave as they approach infinity. Imagine a sequence as a list of numbers that follows a specific rule. We're interested in what happens to these numbers as we progress further and further along the sequence. This is where the limit comes into play.

For any given sequence \( a_n \), if the numbers get closer and closer to a specific value \( L \) as \( n \) becomes very large, the sequence is said to converge to \( L \). Mathematically, this is written as \( a_n \to L \) as \( n\to\infty \).

In the provided exercise, we worked with a sequence \( a_n = \frac{3}{2} + \frac{3}{2n} \). As \( n \to \infty \), the term \( \frac{3}{2n} \) becomes negligible because the denominator is increasing indefinitely, leading to the limit \( a_n \to \frac{3}{2} \).

Triangular Numbers

Triangular numbers form a sequence of numbers derived from a triangular pattern of evenly spaced dots. The nth triangular number is found by arranging dots to form an equilateral triangle and then counting them.

The formula for the nth triangular number is \( T_n = \frac{n(n+1)}{2} \). This simple expression arises because it sums the first \( n \) natural numbers.

In our exercise, triangular numbers appear in the sequence expression: \( \frac{n(n+1)}{2} \). These numbers play a key role in simplifying the sequence to a manageable form so we can easily assess its behavior as \( n \) grows.

Asymptotic Behavior

Asymptotic behavior describes how a function behaves as its input approaches a certain point or infinity. It's essentially about identifying the trend that a sequence or function follows for very large values. This helps in understanding limits and predicting long-term trends without calculating every single term.

When analyzing the sequence \( a_n = \frac{3}{2} + \frac{3}{2n} \), its asymptotic behavior tells us that for very large \( n \), the term \( \frac{3}{2n} \) diminishes and the sequence behaves similarly to a constant: \( \frac{3}{2} \). This is because as \( n \to \infty \), \( \frac{3}{2n} \) becomes insignificant, highlighting the sequence's convergence to its limit.

Infinite Series

An infinite series is the sum of the terms of an infinite sequence. While our exercise doesn't directly involve an infinite series, understanding series is crucial for grasping more complex scenarios involving sequence limits.

An infinite series usually takes the form \( S = a_1 + a_2 + a_3 + \, ... \) which can often converge to a particular value, analogous to sequence convergence. The challenge lies in determining if the series converges and what value it converges to.

Even though we don't calculate an infinite series in this exercise, the convergence concept from sequences holds significant importance in series too, especially when dealing with terms that become negligible, as seen with \( \frac{3}{2n} \) in sequence limits.