Problem 13 - Calculus: Early Transcendentals

Question

Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 1}^{\infty} \frac {10^n}{(n + 1)4^{2n +1}} \)

Step-by-Step Solution

Answer

The series is convergent by the Ratio Test.

Step 1: Identify the terms

Let's denote the nth term of the series as \( a_n = \frac{10^n}{(n + 1)4^{2n + 1}} \). Our goal is to apply the Ratio Test to determine if the series converges.

Step 2: Apply the Ratio Test Formula

The Ratio Test involves calculating \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If this limit is less than 1, the series converges; if greater than 1, it diverges; if equal to 1, the test is inconclusive.

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

Calculate the \((n+1)\)th term: \( a_{n+1} = \frac{10^{n+1}}{(n+2)4^{2(n+1) + 1}} = \frac{10^{n+1}}{(n+2)4^{2n + 3}} \).

Step 4: Set Up the Ratio

Plug \( a_{n+1} \) and \( a_n \) into the ratio \( \frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+2)4^{2n+3}} \times \frac{(n+1)4^{2n+1}}{10^n} \). Simplify this expression.

Step 5: Simplify the Ratio

Simplifying the expression gives us: \[\frac{a_{n+1}}{a_n} = \frac{10^{n+1}(n+1)4^{2n+1}}{10^n(n+2)4^{2n+3}} = \frac{10 \times (n+1)}{(n+2) \times 4^2}\] Thus, \[\frac{a_{n+1}}{a_n} = \frac{10(n+1)}{16(n+2)}\]

Step 6: Find the Limit

Now, take the limit as \( n \to \infty \):\[\lim_{n \to \infty} \left| \frac{10(n+1)}{16(n+2)} \right| = \lim_{n \to \infty} \frac{10n + 10}{16n + 32}\]Factor \( n \) out of the numerator and denominator:\[\lim_{n \to \infty} \frac{10(n/n) + 10/n}{16(n/n) + 32/n} = \frac{10}{16} = \frac{5}{8}\]

Step 7: Conclude by the Ratio Test

Since \( \frac{5}{8} < 1 \), the Ratio Test tells us that the series \( \sum_{n=1}^{\infty} \frac{10^n}{(n+1)4^{2n+1}} \) is convergent.

Key Concepts

Convergence of Series

In mathematics, determining whether a series converges or diverges is crucial for understanding its behavior. When a series converges, it means that the sum of its infinite terms approaches a specific number. Conversely, if a series diverges, no finite value limits the sum of its terms.

The Ratio Test is a popular method for testing convergence of series. By examining the ratio of successive terms, we can decide if the series settles into a constant sum or continues to grow indefinitely. In our given series, we identified the terms and computed the ratio of consecutive terms. Calculating the limit of this ratio as the term index approaches infinity provides insight into the nature of the series.

In practice, if the limit of the absolute value of the ratio is less than 1, the series converges. If greater than 1, the series diverges, and if it equals 1, the test provides no information, and another method must be employed.

Infinite Series

An infinite series is an expression composed of infinitely many terms that follow a specific rule. This rule typically involves consecutive natural numbers as indices. Examples include arithmetic series, geometric series, and more complex forms like the one in our exercise.

Infinite series often appear in calculus, where they aid in approximating functions and solving differential equations. Despite containing infinite terms, some series like the geometric series have finite sums. This attribute hinges on their convergence.

Understanding infinite series is important in calculus and mathematical analysis. They offer tools for modeling real-world phenomena, from computing interest rates in finance to predicting population growth in biology.

Calculus Problem Solving

Solving calculus problems frequently involves examining series, limits, and sequences. Calculus provides powerful tools for handling continuous changes, making it indispensable in fields like physics, engineering, and economics.

In tackling such problems, it’s important to have a methodical approach:

• First, identify the problem’s nature and relevant concepts, such as differentiation, integration, or series manipulation.
• Next, break down the problem into manageable parts, addressing each with appropriate calculus techniques.
• Finally, verify solutions for accuracy and interpret their real-world implications.

Developing strong problem-solving skills in calculus is essential for making sense of complex mathematical models and deriving practical solutions.