Problem 40 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{2^{n}+3^{n}}{3^{n}+4^{n}}\end{equation}

Step-by-Step Solution

Answer

The series converges.

Step 1: Analyze the Growth of the Numerator and Denominator

Consider the terms in the series \(a_n = \frac{2^n + 3^n}{3^n + 4^n}\). As \(n\) becomes very large, notice that \(3^n\) dominates \(2^n\) in the numerator and \(4^n\) dominates \(3^n\) in the denominator. Thus, the expression simplifies to \(a_n \sim \frac{3^n}{4^n} = \left(\frac{3}{4}\right)^n\).

Step 2: Determine the Convergence of the Simplified Series

Since \( a_n \sim \left(\frac{3}{4}\right)^n \), consider the geometric series \( \sum_{n=1}^{\infty} \left(\frac{3}{4}\right)^n \). The common ratio \( r = \frac{3}{4} \) is between \(-1\) and \(1\), indicating convergence of the geometric series.

Step 3: Apply the Limit Comparison Test

To confirm the behavior of the original series, apply the limit comparison test. Consider the series \( b_n = \left(\frac{3}{4}\right)^n \). Evaluate the limit:\[\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{2^n+3^n}{3^n+4^n}}{\left(\frac{3}{4}\right)^n} = \lim_{n \to \infty} \frac{2^n + 3^n}{3^n + 4^n} \cdot \frac{4^n}{3^n}\]Simplifying further yields:\[\lim_{n \to \infty} \frac{4^n(2^n + 3^n)}{3^n(3^n + 4^n)} = \lim_{n \to \infty} \frac{4^n(3^n)}{3^n(4^n)} = 1\]Since this limit is a finite non-zero number, and \( b_n \) is a convergent series, \( a_n \) also converges.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the \emph{common ratio}, denoted by \(r\). The series takes the form:

• \(a, ar, ar^2, ar^3, \ldots\)

The sum of an infinite geometric series converges when the absolute value of the common ratio \(|r|\) is less than 1, leading to the sum:\[S = \frac{a}{1 - r}\]In our exercise, after simplifying the expression, the series \(\sum \left(\frac{3}{4}\right)^n\) was identified as a geometric series with a common ratio of \(\frac{3}{4}\). Since \(|\frac{3}{4}| < 1\), this series converges.Understanding a geometric series is essential because it allows students to quickly determine whether a series converges or diverges based solely on the value of the common ratio.

Limit Comparison Test

The Limit Comparison Test is a useful tool for determining the convergence or divergence of series, especially when the terms of a series are similar to those of a known simpler series. Here's how it works:

• Select two series \(a_n\) and \(b_n\) where \(b_n\) is a series with known convergence behavior.
• Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = L \).
• If the limit \(L\) is a finite, positive number, then both series \(\sum a_n\) and \(\sum b_n\) share the same convergent or divergent behavior.

In the original exercise, the series associated with \(a_n\) was compared to a geometric series \(b_n = \left(\frac{3}{4}\right)^n\). We found that:\[\lim_{n \to \infty} \frac{a_n}{b_n} = 1\]This result is a finite non-zero value, confirming that \(\sum a_n\) converges since \(\sum b_n\) is known to converge. The Limit Comparison Test is particularly powerful because it circumvents more complex analyses by comparing to simpler series.

Convergent and Divergent Series

Understanding the concepts of convergent and divergent series is crucial for series analysis. In simple terms:

• A series is called \emph{convergent} if the sum of its terms approaches a finite number as more terms are added. Mathematically, \(\sum_{n=1}^{\infty} a_n = L\), where \(L\) is a real number.
• A series is \emph{divergent} if its terms do not approach a particular value, which can lead to either infinity or no particular limit.

Convergence is important because convergent series have meaningful sums, whereas divergent series do not. In our example, we showed that the original series \(\sum_{n=1}^{\infty} \frac{2^n+3^n}{3^n+4^n}\) is convergent by employing simplification techniques and tests like the Limit Comparison Test. Recognizing whether a series converges or diverges can affect numerous applications in mathematics and real-world problems, such as evaluating functions, predicting trends, or optimizing calculations.