Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

An explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=(2 n)^{1 / 2 n} $$

Step-by-Step Solution

Answer

The sequence converges, and \(\lim_{n \rightarrow \infty} a_{n} = 1\).

Step 1: Calculate the First Five Terms

To find the first five terms, substitute \(n = 1, 2, 3, 4,\) and \(5\) into the formula \(a_{n} = (2n)^{1/2n}\). - For \(n = 1\): \(a_{1} = (2 \times 1)^{1/(2 \times 1)} = 2^{1/2} = \sqrt{2}\).- For \(n = 2\): \(a_{2} = (2 \times 2)^{1/(2 \times 2)} = 4^{1/4} = \sqrt[4]{4}\).- For \(n = 3\): \(a_{3} = (2 \times 3)^{1/(2 \times 3)} = 6^{1/6}\).- For \(n = 4\): \(a_{4} = (2 \times 4)^{1/(2 \times 4)} = 8^{1/8}\).- For \(n = 5\): \(a_{5} = (2 \times 5)^{1/(2 \times 5)} = 10^{1/10}\).Thus, the first five terms are: \(\sqrt{2}\), \(\sqrt[4]{4}\), \(6^{1/6}\), \(8^{1/8}\), and \(10^{1/10}\).

Step 2: Determine Convergence or Divergence

To determine if the sequence converges or diverges, evaluate the behavior of \(a_{n}\) as \(n\) approaches infinity.The expression \((2n)^{1/2n}\) can be rewritten using exponential notation as \(\exp\left(\frac{1}{2n} \ln(2n)\right)\). As \(n \rightarrow \infty\), \(\ln(2n)\rightarrow \infty\), but \(\frac{1}{2n}\rightarrow 0\).Thus, the exponent \(\frac{1}{2n} \ln(2n)\to \frac{0}{\infty} = 0\), making \(\exp(0) = 1\).Therefore, the sequence converges to \(1\).

Step 3: Find the Limit

Since the sequence converges, find the limit:\[ \lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} (2n)^{1/2n} = 1. \]Hence, the limit of the sequence is \(1\).

Key Concepts

Explicit Formula

An explicit formula in mathematics is a direct expression that allows you to compute any term in a sequence without referring to previous ones. For sequences, it's a kind of shortcut that bypasses the need for calculating all previous terms. This can be particularly useful when you need to find terms that are far along in the sequence.

In this exercise, the explicit formula provided is \(a_{n} = (2n)^{1/2n}\). This means that for any positive integer \(n\), the \(n\)-th term of our sequence is given by \((2n)^{1/2n}\).

The formula involves basic operations with the variable \(n\), demonstrating a powerful aspect of explicit formulas, their ability to generalize terms based on a specific pattern. Using explicit formulas, you can easily generate the first few terms, or any specific term, as shown in the step-by-step solution where \(n = 1, 2, 3, 4,\) and \(5\) to discover the first five terms.

Limit of a Sequence

The limit of a sequence is what the terms of the sequence approach as \(n\) becomes very large, potentially infinity. It's a central concept in calculus and helps in determining the behavior of sequences over long terms.

When we talk about a sequence converging, we mean that as \(n\) increases towards infinity, the terms get closer and closer to a particular value, which we refer to as the limit of the sequence. If no such value exists, we say the sequence diverges.

In the given problem, the sequence \((2n)^{1/2n}\) converges to \(1\). This was determined by taking the limit as \(n\) approaches infinity. By rewriting the expression as an exponential form \(\exp(\frac{1}{2n} \ln(2n))\), and observing that \(\frac{1}{2n} \rightarrow 0\) as \(n\) increases, we see that the whole expression \(\rightarrow 0\), thus making \(\exp(0) = 1\). Hence, the limit of the sequence as \(n\) approaches infinity is \(1\).

First Five Terms of a Sequence

The first five terms of a sequence provide a foundational understanding of the pattern and growth manner of that sequence. It involves substituting the first few natural numbers into the explicit formula of the sequence.

In our example, to find the first five terms using \(a_{n} = (2n)^{1/2n}\), we substitute values:

• For \(n = 1\): \(a_{1} = \sqrt{2}\)
• For \(n = 2\): \(a_{2} = \sqrt[4]{4}\)
• For \(n = 3\): \(a_{3} = 6^{1/6}\)
• For \(n = 4\): \(a_{4} = 8^{1/8}\)
• For \(n = 5\): \(a_{5} = 10^{1/10}\)

These terms reflect the pattern derived from the formula and show how each term fits into the overall sequence calculation. Once computed, these terms can be used to gain insights into the sequence's behavior, such as its tendency towards convergence or divergence.