Problem 92 - Precalculus with Limits

Question

In Exercises 85-96, find the sum. \( \displaystyle \sum_{j=3}^{5} \frac{1}{j^2 - 3} \)

Step-by-Step Solution

Answer

The sum is \(\frac{1}{6} + \frac{1}{13} + \frac{1}{22}\)

Step 1: Identify Components

The general form of this summation is \(\frac{1}{j^2 - 3}\) and the lower and upper bounds are 3 and 5, respectively.

Step 2: Substitute Values

For every value of j from 3 to 5, substitute it into \(\frac{1}{j^2 - 3}\): For \(j = 3\), the term is \(\frac{1}{3^2 - 3} = \frac{1}{6}\). For \(j = 4\), the term is \(\frac{1}{4^2 - 3} = \frac{1}{13}\). For \(j = 5\), the term is \(\frac{1}{5^2 - 3} = \frac{1}{22}\).

Step 3: Adding-up

Adding those terms, the final sum is therefore \(\frac{1}{6} + \frac{1}{13} + \frac{1}{22}\)

Key Concepts

Fraction Simplification

In the given problem, the expression inside the summation is a fraction: \( \frac{1}{j^2 - 3} \). To successfully simplify this fraction, it's essential to understand what is being expressed here. We are essentially finding the reciprocal of the expression \( j^2 - 3 \). Each time you substitute a new value of \( j \) into this expression, a new fraction is formed. Here's a closer look at simplifying these fractions:

• Analyze the Denominator: The denominator is a quadratic expression, \( j^2 - 3 \). It varies depending on the value of \( j \) you substitute into it.
• Evaluate for Each \( j \): When \( j = 3 \), the fraction simplifies to \( \frac{1}{6} \) because \( 3^2 - 3 = 6 \). Whenever possible, break down terms to their simplest form. This will make further calculations easier.
• Repeat for Each Term: Repeating this process for \( j = 4 \) and \( j = 5 \) gives \( \frac{1}{13} \) and \( \frac{1}{22} \) respectively.

Simplification is all about reducing fractions to their simplest form, ensuring all calculations that follow remain as straightforward as possible.

Series Evaluation

Series evaluation refers to the process of finding the sum of the terms of a sequence. In the exercise, we deal with a small finite series defined by the summation notation \( \sum_{j=3}^{5} \frac{1}{j^2 - 3} \). To evaluate this series, we aim to add the individual terms derived from each value of \( j \) within the defined bounds.Let's break down the steps to evaluate this type of series:

• Define the Terms: First, understand that each term of this series is an individual fraction calculated by substituting each integer from 3 to 5 into \( \frac{1}{j^2 - 3} \).
• Add the Fractions: Once these terms are calculated (\( \frac{1}{6}, \frac{1}{13}, \frac{1}{22} \)), they are summed together to find the series' total value.
• Simplify the Result: Adding fractions involves finding a common denominator or simply adding them using a calculator for precision.

Once the fractions are added, the sum \( \frac{1}{6} + \frac{1}{13} + \frac{1}{22} \) gives you the result of this series evaluation.

Mathematical Sequences

Mathematical sequences are ordered lists of numbers defined by a specific rule. In this exercise, the sequence involves integer values from 3 to 5, each providing a term within the series \( \sum_{j=3}^{5} \frac{1}{j^2 - 3} \).To better understand sequences, consider:

• Sequence Definition: A sequence is defined by a starting point and a rule for determining subsequent terms. Here, the rule is given by \( \frac{1}{j^2 - 3} \) and the sequence values are the integers from 3 to 5.
• Finite versus Infinite: The sequence in this exercise is finite because it ends after three terms. Sequences can also be infinite, depending on the problem context.
• Applications: Identifying a sequence helps break down complex problems into smaller, manageable parts.

By analyzing a mathematical sequence, it's easier to comprehend and solve summation problems like the given exercise, leading to efficient series evaluations and overall understanding.