Problem 41 - Precalculus with Limits

Question

In Exercises 39-48, write the first five terms of the sequence and find the limit of the sequence (if it exists). If the limit does not exist, explain why. Assume \(n\) begins with 1. $$ a_n = \dfrac{n}{2n+1} $$

Step-by-Step Solution

Answer

The first five terms of the sequence are \(\frac{1}{3}, \frac{2}{5}, \frac{3}{7}, \frac{4}{9}, \frac{5}{11}\) and the limit of the sequence is \(\frac{1}{2}\).

Step 1: Find the first five terms of the sequence

To find the first five terms of the sequence \(a_n = \frac{n}{2n+1}\), substitute the values of \(n\) starting from 1 up to 5. The five terms are obtained as follows: \n \(a_1 = \frac{1}{2*1+1} = \frac{1}{3}\) \n \(a_2 = \frac{2}{2*2+1} = \frac{2}{5}\) \n \(a_3 = \frac{3}{2*3+1} = \frac{3}{7}\) \n \(a_4 = \frac{4}{2*4+1} = \frac{4}{9}\) \n \(a_5 = \frac{5}{2*5+1} = \frac{5}{11}\)

Step 2: Find the limit of the sequence

The limit of the sequence can be found by taking the limit of the sequence as \(n\) approaches infinity. We thus find: \n \(\lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2} \) To see why, we can divide the numerator and the denominator by n (the highest power in the equation) which gives us this: \n \(\lim_{n\to\infty} \frac{1}{2 + \frac{1}{n}} = \frac{1}{2}\) As \(n\) goes to infinity, \(\frac{1}{n}\) goes to 0, thus the limit of the sequence is \(\frac{1}{2}\).

Key Concepts

Mathematical Sequences

Mathematical sequences are ordered lists of numbers that follow a particular rule or pattern. The individual elements in the sequence are usually called terms. Sequences can be finite, with a set number of terms, or infinite, continuing indefinitely. In our given exercise, the sequence is presented by the formula \( a_n = \frac{n}{2n+1} \).

Understanding sequences is foundational for various branches of mathematics, including calculus and algebra. The terms of the sequence are often represented using the variable \(n\), which typically starts at 1 and increases by 1 with each subsequent term. This index \(n\) indicates the term's position within the sequence.

Finding Sequence Terms

To find individual terms of a sequence, simply plug the position number (n) into the formula that defines the sequence. For the sequence \( a_n = \frac{n}{2n+1} \), the terms are found by substituting \(n\) with consecutive integers starting from 1. Here's how you can calculate the first five terms:

• For \(n=1\), \(a_1 = \frac{1}{3}\)
• For \(n=2\), \(a_2 = \frac{2}{5}\)
• For \(n=3\), \(a_3 = \frac{3}{7}\)
• For \(n=4\), \(a_4 = \frac{4}{9}\)
• For \(n=5\), \(a_5 = \frac{5}{11}\)

Each term requires a simple calculation. This direct computation method can be applied to any explicitly defined sequence.

Convergence of Sequences

A key concept in calculus is the convergence of sequences. A sequence converges if, as \(n\) becomes very large, the terms of the sequence approach a specific value, called the limit. If a sequence doesn't have such a limit, it is said to diverge.

Looking back at our example, \( a_n = \frac{n}{2n+1} \), we aim to find out if it converges and to what value. Determining convergence usually involves taking the limit as \(n\) approaches infinity. If the ratio of consecutive terms stabilizes or approaches a fixed number, this number is the limit of the sequence, indicating convergence. If the terms keep growing without bounds or oscillate indefinitely without settling, the sequence diverges.

Limits at Infinity

Limits at infinity are involved in assessing the behavior of sequences or functions as their input (or index, for sequences) grows without bounds. For sequences, it reflects what value the terms approach as you move far out in the sequence.

The limit for the sequence in the exercise is calculated by dividing numerator and denominator by \(n\), the highest power in the equation, leading to: \[\frac{1}{2+\frac{1}{n}}\]. As \(n\) goes to infinity, the fraction \(\frac{1}{n}\) approaches zero, which simplifies the expression to \( \frac{1}{2} \), making the limit of the sequence \( \frac{1}{2} \). This limit shows that as \(n\) increases, the value of each term gets closer and closer to \( \frac{1}{2} \), concluding that the sequence converges to this value.