Problem 44 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{(2 n+3)\left(2^{n}+3\right)}{3^{n}+2} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Identify the Series

The series in question is \( \sum_{n=1}^{\infty} \frac{(2n+3)(2^n+3)}{3^n+2} \). We need to determine if this series converges or diverges.

Step 2: Apply the Ratio Test

The Ratio Test can be used here. Define \( a_n = \frac{(2n+3)(2^n+3)}{3^n+2} \) and compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

Step 3: Calculate \( a_{n+1} \)

Calculate \( a_{n+1} = \frac{(2(n+1)+3)(2^{n+1}+3)}{3^{n+1}+2} \). Simplify it to \( \frac{(2n+5)(2 \cdot 2^n + 3)}{3 \cdot 3^n + 2} \).

Step 4: Compute \( \frac{a_{n+1}}{a_n} \)

Find \( \frac{a_{n+1}}{a_n} = \frac{(2n+5)(2 \cdot 2^n + 3)(3^n + 2)}{(2n+3)(2^n+3)(3^{n+1} + 2)} \). Simplify this expression.

Step 5: Simplify the Expression

Upon simplification, \( \frac{a_{n+1}}{a_n} = \frac{(2n+5)(2^{n+1} + 3)(3^n + 2)}{(2n+3)(2^n + 3)(3 \cdot 3^n + 2)} \approx \frac{2}{3} \) for large \( n \).

Step 6: Evaluate the Limit

Taking the limit as \( n \) approaches infinity, we get \( L = \lim_{n \to \infty} \frac{2}{3} = \frac{2}{3} \).

Step 7: Conclusion from the Ratio Test

Since \( L = \frac{2}{3} < 1 \), according to the Ratio Test, the series \( \sum_{n=1}^{\infty} \frac{(2n+3)(2^n+3)}{3^n+2} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a useful tool to determine if an infinite series converges or diverges. It focuses on the ratio of successive terms in a series. Essentially, you compute the limit

• Define the terms: Suppose you have a series with terms noted as \( a_n \).
• Find the next term's ratio: Compute the absolute value of the ratio between the \( (n+1)^{th} \) term and the \( n^{th} \) term: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• Limit determines convergence: If this limit \( L < 1 \), the series converges. If \( L > 1 \) or is infinite, the series diverges. If \( L = 1 \), the test is inconclusive.

In our example, after calculations, \( L = \frac{2}{3} \), which is less than 1. According to the Ratio Test, this means that our series converges.

Infinite Series

An infinite series is simply the sum of an infinite sequence of terms. In notations, it’s expressed as \( \sum_{n=1}^{\infty} a_n \). These series can either converge or diverge, based on the values of their terms and the cumulative effect of these terms being added together.

• Converging Series: When the sum of the series approaches a finite number as the number of terms grows to infinity.
• Diverging Series: When the sum of the series doesn’t settle to a number and continues to increase without bound or oscillates indefinitely.

Understanding convergence and divergence is crucial for working with infinite series in calculus and helps in evaluating functions and complex calculations.

Convergence and Divergence

The concept of convergence and divergence is fundamental in calculus when dealing with infinite series. Determining whether a series converges or diverges tells us if it adds up to a finite number.

• Convergence: If a series' partial sums approach a specific value, the series converges. This means that as you keep adding more terms, the total sum gets closer to a particular number.
• Divergence: If the partial sums do not approach a specific value, the series diverges. The sum could either grow indefinitely or cycle without settling.

In practice, tests like the Ratio Test, Integral Test, or Comparison Test are employed to assess the behavior of a series.

Limit Comparison

The Limit Comparison Test is another method used to analyze series' behavior. It’s especially helpful when the series terms look complicated or when other tests aren’t easily applicable.

• Choose a comparison series \( b_n \) that is similar to \( a_n \) and whose convergence property is known.
• Compute the limit: \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If the limit is a positive finite number, both series \( \sum a_n \) and \( \sum b_n \) either both converge or both diverge.

This technique relies on the observation that if two series' terms grow at a similar rate, they likely share the same convergence properties. Though not used in this particular solution, the Limit Comparison Test is a versatile tool worth knowing.