Question

Evaluate. $$ \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} $$

Short answer

The sum is \( \frac{1}{2} \).

Step-by-step solution

Understanding the Series

The series given is \( \sum_{n=1}^{\infty} \frac{1}{(n+1)(n+2)} \). This is an infinite series, and it seems to involve partial fractions, where each term looks like a fraction with polynomial factors in the denominator.

Applying Partial Fraction Decomposition

Rewrite the term \( \frac{1}{(n+1)(n+2)} \) as a sum of simpler fractions using partial fraction decomposition. Assume \( \frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2} \). Solve for \( A \) and \( B \) by clearing the denominators: \( 1 = A(n+2) + B(n+1) \). After equating coefficients, find that \( A = 1 \) and \( B = -1 \). Thus, \( \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2} \).

Writing the Infinite Series

Now that the partial fraction decomposition is found, substitute it back into the series: \( \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \). Notice that this can lead to a telescoping series.

Understanding Telescoping Series

A telescoping series is one where most terms cancel each other out. Write the first few terms explicitly to see the pattern: \( \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \ldots \). In this series, each term \(-\frac{1}{n+2}\) cancels with \(\frac{1}{n+2}\) in the next term.

Calculating the Sum

With a telescoping series, mainly the first term's part and the last term remain. Hence, the series effectively reduces to \(\frac{1}{2}\). As \( n \to \infty \), all other terms cancel out and you are left with just \(\frac{1}{2} \).

Final Answer

The evaluated sum of the infinite series is \( \frac{1}{2} \).

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in calculus to break down complex rational expressions into simpler fractions. This helps simplify integration and summation tasks. In the context of an infinite series, it allows us to transform a complicated term into something more manageable. For example, consider the expression \( \frac{1}{(n+1)(n+2)} \). The goal is to express it in terms of simpler fractions that can be summed easily. By assuming \( \frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2} \), we equate the two sides and solve for \( A \) and \( B \). By clearing denominators and matching coefficients, we find \( A = 1 \) and \( B = -1 \). This leads to the expression:\[ \frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2} \]This approach simplifies the original complex fraction by expressing it in terms of simpler parts that are easier to manage in calculations.

Telescoping Series

A telescoping series is a type of series in which many terms cancel out when summed, making it much simpler to evaluate. This concept comes in handy when dealing with series whose terms are expressed as the difference of consecutive terms. In the given exercise, the series becomes:\[ \sum_{n=1}^{\infty} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \]Once expanded, the series looks like:

• \( \frac{1}{2} - \frac{1}{3} \)
• \( \frac{1}{3} - \frac{1}{4} \)
• \( \frac{1}{4} - \frac{1}{5} \)
• And so on...

In this pattern, each term \(-\frac{1}{n+2}\) cancels with the \(\frac{1}{n+2}\) term in the subsequent fraction. This cancellation effect removes most terms, leaving only the first positive term and the last negative term in finite telescoping, and remarkably simplifying the infinite series to a single value.

Calculus

Calculus is a branch of mathematics that focuses on continuous change. It plays a critical role in evaluating infinite series because it provides the tools needed to understand and calculate sums and limits effectively. With an infinite series, like in our problem, calculus helps us recognize patterns that emerge when summing an infinite number of terms.The tools of calculus, such as limits and series convergence, allow us to evaluate expressions where traditional arithmetic might not suffice. When working with infinite series, we often look at how terms behave as \( n \to \infty \). In the case of a telescoping series, calculus helps us conclude that the infinite cancellations leave us with a finite sum. This understanding comes from calculus principles that underpin series evaluation. Ultimately, the final evaluated sum of an infinite series like the one in the exercise is achieved through evaluating limits and effectively using simplification techniques.