Problem 63 - Precalculus: Mathematical for Calculus

Question

Write the sum using sigma notation. $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{999 \cdot 1000}$$

Step-by-Step Solution

Answer

\(\sum_{n=1}^{999} \frac{1}{n(n+1)}\)

Step 1: Identify Pattern

Look at the fractions given and notice that each fraction uses consecutive integers in its denominator. The first fraction has a denominator of 1 and 2, the next has 2 and 3, and so on until 999 and 1000.

Step 2: General Formula for the nth Term

Express the general form for the nth term of the sequence. The numerator is always 1, and the denominator for the nth term is the product of two consecutive integers: \[\frac{1}{n(n+1)}\] where \(n\) starts at 1.

Step 3: Determine the Range of n

Given the last term is \(\frac{1}{999 \cdot 1000}\), set the final pair of numbers in the sequence equal to the general formula's denominator terms. This means \(n+1 = 1000\), which gives \(n = 999\). The sequence starts with \(n = 1\).

Step 4: Write in Sigma Notation

Combine the information to express the sum in sigma notation:\[\sum_{n=1}^{999} \frac{1}{n(n+1)}\] This notation compactly represents the sum of fractions from the first term to the term where \(n=999\).

Key Concepts

Sequence

Sequences are ordered lists of numbers that follow a specific pattern or rule. In this exercise, the sequence consists of fractions where each fraction has a numerator of 1 and a denominator that is the product of two consecutive integers.
For example:

• The first fraction is \( \frac{1}{1 \cdot 2} \)
• The second fraction is \( \frac{1}{2 \cdot 3} \)
• The third fraction is \( \frac{1}{3 \cdot 4} \)

This pattern continues up to the 999th term in the sequence, which is \( \frac{1}{999 \cdot 1000} \).
Identifying the pattern in a sequence is crucial for creating a general formula that can describe any term within the sequence. Here, each term \( a_n \) is represented by the formula \( \frac{1}{n(n+1)} \), where \( n \) is an integer starting from 1.
Sequences can be finite or infinite, depending on the range of terms you are working with. In this case, our sequence is finite because it ends at \( n = 999 \). Understanding sequences is essential for recognizing how terms are built over intervals. This is necessary for further operations such as summation using sigma notation.

Series

A series is the sum of terms in a sequence. When you see a plus sign adding each term in our sequence, like \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{999 \cdot 1000} \), you're dealing with a series.
In this exercise, we worked on expressing this series using sigma notation, which is a compact form that summarizes long sums. It is written as:
\[ \sum_{n=1}^{999} \frac{1}{n(n+1)} \]
This notation indicates that you are summing the function \( \frac{1}{n(n+1)} \) for every integer \( n \) starting at 1 and ending at 999. Sigma notation simplifies the writing and calculation of series, making it easier to handle complex operations.
Understanding how to convert a series into sigma notation is vital for mathematical communication. It translates lengthy expressions into simpler, more efficient forms that are easier to analyze and compute, especially when dealing with large numbers of terms.

Fractions

Fractions are numerical expressions representing the division of one quantity by another. In our exercise, each term in the sequence is a fraction where the numerator is consistently 1, and the denominator varies based on consecutive integers.
For example, the denominator of each fraction in this sequence takes the form \( n(n+1) \), providing the structure for each term's specific position.
This way:

• The denominator of the first term is \( 1 \cdot 2 \)
• The second term is \( 2 \cdot 3 \)
• This pattern continues up to \( 999 \cdot 1000 \)

Understanding how to manipulate fractions is fundamental in algebra and calculus as it allows you to simplify expressions and solve equations more efficiently.
Firstly, recognize the numerator and denominator's roles in a fraction. In particular sequences or series, the denominator often reflects the pattern followed by the sequence. Knowing this helps in forming general formulas like \( \frac{1}{n(n+1)} \) for sequences and series operations.