Problem 47 - Thomas Calculus

Question

Find the sum of each series in Exercises \(41-48\) $$ \sum_{n=1}^{\infty}\left(\frac{1}{\ln (n+2)}-\frac{1}{\ln (n+1)}\right) $$

Step-by-Step Solution

Answer

The sum of the series is \(-\frac{1}{\ln 2}\).

Step 1: Identify the Type of Series

The given series is telescoping, which can be identified by the format where terms cancel out in a specific pattern. The series is written as \( \sum_{n=1}^{\infty} \left( \frac{1}{\ln (n+2)} - \frac{1}{\ln (n+1)} \right) \).

Step 2: Write Out the First Few Terms

To understand the pattern in which terms cancel, write out the first few terms of the series: \( \left( \frac{1}{\ln 3} - \frac{1}{\ln 2} \right) + \left( \frac{1}{\ln 4} - \frac{1}{\ln 3} \right) + \left( \frac{1}{\ln 5} - \frac{1}{\ln 4} \right) + \ldots \).

Step 3: Observe Cancellation Pattern

Notice the pattern where the \( \frac{1}{\ln (n+2)} \) term of one expression cancels with the \( \frac{1}{\ln (n+1)} \) term of the subsequent expression. Each positive term \( \frac{1}{\ln (n+2)} \) cancels with the negative term in the next parenthesis.

Step 4: Simplify the Series

Upon canceling all intermediate terms, the series simplifies to the first negative term and the limit of the positive term as \( n \to \infty \). The simplified expression is: \( - \frac{1}{\ln 2} + \lim_{n \to \infty} \frac{1}{\ln (n+2)} \).

Step 5: Evaluate the Limit

Evaluate \( \lim_{n \to \infty} \frac{1}{\ln (n+2)} \), which approaches zero because as \( n \to \infty \), \( \ln (n+2) \) increases indefinitely, making its reciprocal vanish.

Step 6: Sum of the Series

By combining the results, the sum of the series is \( - \frac{1}{\ln 2} + 0 = - \frac{1}{\ln 2} \).

Key Concepts

Infinite Series

An infinite series is a sum of an infinite sequence of terms. When dealing with series, we are looking to find the total sum as more and more terms are added. The fascinating aspect of infinite series is that despite having an infinite number of terms, they can still converge to a finite value. This process involves understanding how terms behave as they progress towards infinity. In the context of the given exercise, the series provided is an example of what's called a telescoping series, where subsequent terms cancel out each other

• Infinite series can have positive or negative terms.
• They can converge (approach a finite value) or diverge (grow indefinitely).
• Different methods, such as telescoping or geometric series, are used to evaluate them.

Studying infinite series is crucial in calculus and analysis as they appear frequently in mathematical modeling and scientific computations.

Logarithms

Logarithms are the inverse operations of exponentials, and they help us solve equations involving powers of numbers. In essence, the logarithm of a number is the exponent to which the base must be raised to produce that number. For this exercise, natural logarithms, denoted as \( \ln(x) \), are primarily used. They use the base \( e \), an irrational and transcendental number approximately equal to 2.718.

• Logarithms reduce complex multiplication into simpler addition.
• Natural logarithms are crucial in calculus for integrating exponential functions.
• They are prominently used in many fields such as science, engineering, and finance.

In the given series, logarithms are key since each term is structured around these functions, impacting how terms cancel when telescoping occurs.

Limits

Limits deal with the behavior of a function as it approaches a particular input or point. They are fundamental in calculus and underpin the definition of derivatives and integrals. In our exercise, limits determine the value that a function approaches as the index \( n \) heads towards infinity. It's important because infinite series require analyzing terms as they progress towards infinity.

• The notation \( \lim_{n \to \infty} \) signifies a limit as \( n \) increases without bound.
• Limits can approach finite numbers, infinity, or incomprehensible values.
• Evaluating limits is crucial for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

In the solution to our series, the limit of \( \frac{1}{\ln (n+2)} \) as \( n \to \infty \) is zero because the logarithm's value increases indefinitely, diminishing its reciprocal.

Convergence of Series

Convergence in series refers to whether the sum of a series approaches a finite value as more terms are added. For a series to converge, its terms must tend towards zero as the series progresses towards infinity. A telescoping series, like the one in the exercise, simplifies this process by having canceling intermediate terms.

• A series converges if the sequence of partial sums approaches a certain limit.
• If a series does not converge, it is said to diverge, leading to an infinite sum.
• Determining convergence involves various tests such as the ratio test, root test, or comparison test.

The telescoping nature of our series simplifies its evaluation. Despite having infinite terms, cancellations lead to a clear remaining value, showing the series' convergence.