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Chapter 5: Problem 57

a. Determine the partial fraction decomposition $$ \text { for } \frac{2}{n(n+2)} . $$ b. Use the partial fraction decomposition for \(\frac{2}{n(n+2)}\) to rewrite the infinite sum $$ \frac{2}{1(3)}+\frac{2}{2(4)}+\frac{2}{3(5)}+\frac{2}{4(6)}+\frac{2}{5(7)} \cdots $$ c. Determine the value of \(\frac{1}{n+2}\) as \(n \rightarrow \infty\). d. Find the value of the sum from part (b).

Short Answer

Expert verified
The sum of the given series is \( \frac{3}{2} \).

Step by step solution

01

Set up the partial fraction decomposition

Given the expression \(\frac{2}{n(n+2)}\), set it up for partial fraction decomposition. \[ \frac{2}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2} \]
02

Find A and B

Combine the fractions on the right-hand side to have a common denominator: \[ \frac{A}{n} + \frac{B}{n+2} = \frac{A(n+2) + Bn}{n(n+2)} = \frac{An + 2A + Bn}{n(n+2)} = \frac{(A + B)n + 2A}{n(n+2)} \] Set the numerators equal to each other: \[ 2 = (A+B)n + 2A \] To solve for A and B, compare the coefficients: \[ A + B = 0 \] \[ 2A = 2 \] Therefore: \[ A = 1 \] \[ B = -1 \]
03

Write the partial fraction decomposition

Now substitute A and B back into the original decomposition: \[ \frac{2}{n(n+2)} = \frac{1}{n} - \frac{1}{n+2} \]
04

Rewrite the infinite sum

Using the partial fraction decomposition, rewrite the given infinite sum: \[ \sum_{n=1}^{\infty} \frac{2}{n(n+2)} = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+2} \right) \]
05

Simplify the sum

Notice that in the series, most of the terms cancel out. For example: \[ \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{2} - \frac{1}{4} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \left( \frac{1}{4} - \frac{1}{6} \right) + \cdots \] This is a telescoping series where most terms cancel out, leaving: \[ 1 + \frac{1}{2} - \left( \lim_{n \to \infty} \frac{1}{n+1} + \frac{1}{n+2} \right) \]
06

Evaluate the limit

As \(n \to \infty\), \(\frac{1}{n+1} \) and \(\frac{1}{n+2} \) approach zero: \[ \lim_{n \to \infty} \frac{1}{n+1} + \frac{1}{n+2} = 0 \]
07

Determine the final value of the sum

Combine the initial terms from the remaining sum: \[ 1 + \frac{1}{2} = \frac{3}{2} \]

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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 kind of series where each term cancels out a part of the previous or next term. This cancellation significantly simplifies the sum. For example, in our series \(\frac{2}{n(n+2)}, \) after partial fraction decomposition, we get \(\frac{1}{n} - \frac{1}{n+2}. \) So, our series becomes:
\[ \frac{1}{1} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \text{...} \]
Notice how most terms cancel out. Now, we only need to look at the uncanceled terms, which simplify our sum evaluation greatly. This is why telescoping series are powerful for simplification.
Infinite Sum
An infinite sum, or infinite series, is the sum of an infinite sequence of numbers. It puts the idea of a series to the extreme, extending forever. In our case, once we decomposed \(\frac{2}{n(n+2)}\), we rewrote our infinite sum:
\[ \frac{2}{1(3)} + \frac{2}{2(4)} + \frac{2}{3(5)} + \frac{2}{4(6)} + \text{...} = \frac{1}{1} - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \frac{1}{3} - \frac{1}{5} + \frac{1}{4} - \frac{1}{6} + \text{...} \]
In infinite sums, understanding the behavior of partial sums is crucial. Here, we're able to simplify this sum using the telescoping property, leaving us with just a few terms to sum up.
Limit as n Approaches Infinity
When we deal with terms like \(\frac{1}{n+1} \) and \(\frac{1}{n+2}\), it's important to understand their behavior as \(n\) grows larger. As \(n\) approaches infinity, these terms approach zero:
\[ \frac{1}{n+1} \to 0 \text{ and } \frac{1}{n+2} \to 0 \text{ as } n \to \text{infinity}\]
This is because the denominator becomes very large, making the fraction very small. Hence, in our telescoping series, these terms vanish in the limit, simplifying the final sum.
Sum Evaluation
Finally, to evaluate our infinite sum, we look at the leftover terms after telescoping and taking the limit. Initially, most terms cancel out, and we are left with:
\[ 1 + \frac{1}{2} - \text{(terms going to zero)} \]
Since \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) go to zero as \(n\) approaches infinity:
\[ 1 + \frac{1}{2} - 0 - 0 = 1.5 \text{ or } \frac{3}{2}\]
This is the value of the sum from part (b).

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Most popular questions from this chapter

Josh makes \(\$ 24 /\) hr tutoring chemistry and \(\$ 20 / \mathrm{hr}\) tutoring math. Let \(x\) represent the number of hours per week he spends tutoring chemistry. Let \(y\) represent the number of hours per week he spends tutoring math. a. Write an objective function representing his weekly income for tutoring \(x\) hours of chemistry and \(y\) hours of math. b. The time that Josh devotes to tutoring is limited by the following constraints. Write a system of inequalities representing the constraints. \- The number of hours spent tutoring each subject cannot be negative. \- Due to the academic demands of his own classes he tutors at most \(18 \mathrm{hr}\) per week. \- The tutoring center requires that he tutors math at least 4 hr per week. \- The demand for math tutors is greater than the demand for chemistry tutors. Therefore, the number of hours he spends tutoring math must be at least twice the number of hours he spends tutoring chemistry. c. Graph the system of inequalities represented by the constraints. d. Find the vertices of the feasible region. e. Test the objective function at each vertex. f. How many hours tutoring math and how many hours tutoring chemistry should Josh work to maximize his income? g. What is the maximum income? h. Explain why Josh's maximum income is found at a point on the line \(x+y=18\).

Explain how to find the solution set to a system of inequalities in two variables.

Solve the system of equations by using the substitution method. (See Example 2\()\) $$ \begin{array}{rr} x+3 y= & 5 \\ 3 x-2 y= & -18 \end{array} $$

The average weekly salary of two employees is \(\$ 1350\). One makes \(\$ 300\) more than the other. Find their salaries

A system of equations is given in which each equation is written in slope- intercept form. Determine the number of solutions. If the system does not have one unique solution, state whether the system is inconsistent or whether the equations are dependent. $$ \begin{array}{l} y=8 x-\frac{1}{2} \\ y=8 x-\frac{1}{2} \end{array} $$
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