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Magazine Chapter 9: Problem 7
Determine whether the series converges, and if so find its sum. $$\sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)}$$
Short Answer
Expert verified
The series converges and its sum is \( \frac{1}{3} \).
Step by step solution
01
Identify the Series Type
The given series is a sum of fractions with polynomial denominators: \( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \). This type of series suggests that we might be able to apply partial fraction decomposition, as it deals with a product of linear factors in the denominator.
02
Decompose Using Partial Fractions
To express \( \frac{1}{(k+2)(k+3)} \) using partial fractions, assume \( \frac{1}{(k+2)(k+3)} = \frac{A}{k+2} + \frac{B}{k+3} \). Solve for \( A \) and \( B \) by equating components of the equation: \( 1 = A(k+3) + B(k+2) \). This results in two equations: \( A + B = 0 \) and \( 3A + 2B = 1 \). Solve these equations to find \( A = 1 \) and \( B = -1 \). Thus, \( \frac{1}{(k+2)(k+3)} = \frac{1}{k+2} - \frac{1}{k+3} \).
03
Rewrite the Series Using the Decomposition
Using the partial fraction decomposition, rewrite the series as \( \sum_{k=1}^{\infty} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \). This series is now a telescoping series, where successive terms cancel each other.
04
Evaluate the Series
To evaluate the telescoping series, write out the first few terms: \( \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \dots \). Notice that each term cancels with the successive term, leaving only the first few terms that do not get cancelled. The series behaves as: \( \frac{1}{3} \) plus a vanishing remainder as \( k \to \infty \).
05
Determine the Sum and Convergence
The series converges because it is a telescoping series where the terms effectively cancel out, except for the very first partial term. The sum of the series is therefore \( \frac{1}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Telescoping Series
A telescoping series is a special type of series where each term cancels out a part of the previous term. This is particularly useful for determining convergence and finding the sum of a series. Let's explore how this works with an example:
The series \( \sum_{k=1}^{\infty} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \) is a classic case of a telescoping series. When expanded, most terms cancel out, significantly simplifying the process of finding the sum.
The beauty of telescoping lies in its ability to reduce complexity by allowing easy calculation of the series sum, often resulting in only a few terms left uncancelled.
The series \( \sum_{k=1}^{\infty} \left( \frac{1}{k+2} - \frac{1}{k+3} \right) \) is a classic case of a telescoping series. When expanded, most terms cancel out, significantly simplifying the process of finding the sum.
- Consider writing out the initial terms: \( \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \dots \)
- Notice how each fraction from one term matches the opposite fraction in the subsequent term, thus canceling each other out.
- What remains is simply the unfollowed initial term, \( \frac{1}{3} \), with the disappearing tail adding nothing as \( k \to \infty \).
The beauty of telescoping lies in its ability to reduce complexity by allowing easy calculation of the series sum, often resulting in only a few terms left uncancelled.
Partial Fractions
Partial fractions are a method used to simplify complex rational expressions, particularly useful in series convergence problems. The technique involves expressing a single complex fraction as a sum of simpler fractions, making the series easier to manage.For our series \( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \), we rewrite the term using partial fractions:
This decomposition transforms the original series into a telescoping series, illustrating how partial fractions simplify the analysis and allow the series to be easily evaluated.
- Express \( \frac{1}{(k+2)(k+3)} \) in terms of \( \frac{A}{k+2} + \frac{B}{k+3} \).
- Solving for constants \( A \) and \( B \) requires creating a single equation: \( 1 = A(k+3) + B(k+2) \).
- This yields two simpler equations: \( A + B = 0 \) and \( 3A + 2B = 1 \).
- Solving these gives \( A = 1 \) and \( B = -1 \), leading to the decomposed form: \( \frac{1}{k+2} - \frac{1}{k+3} \).
This decomposition transforms the original series into a telescoping series, illustrating how partial fractions simplify the analysis and allow the series to be easily evaluated.
Infinite Series
An infinite series is a sum of an infinite sequence of terms. Understanding the convergence of infinite series is key to many mathematical applications.Consider the series \( \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+3)} \), which is an example of an infinite series. The main goal is to determine whether this series converges, and if so, to find the sum.
This convergence implies a finite sum, demonstrating the power and utility of infinite series in calculating sums where terms continuously stack up to a limit.
- A series converges if the sequence of its partial sums approaches a specific value as the number of terms increases indefinitely.
- For our given series, the use of partial fractions leads to a telescoping series, simplifying the sum to just the first term that survives after cancellation, \( \frac{1}{3} \).
- Therefore, by using mathematical techniques like partial fractions and recognizing telescoping patterns, we conclude that the series converges to \( \frac{1}{3} \).
This convergence implies a finite sum, demonstrating the power and utility of infinite series in calculating sums where terms continuously stack up to a limit.
