Problem 41 - Precalculus: Mathematical for Calculus

Question

Find the sum. $$\sum_{k=1}^{3} \frac{1}{k}$$

Step-by-Step Solution

Answer

The sum is \( \frac{11}{6} \).

Step 1: Understand the Summation

The expression \( \sum_{k=1}^{3} \frac{1}{k} \) means that we need to evaluate and add the terms \( \frac{1}{k} \) for each integer value of \( k \) from 1 to 3.

Step 2: Evaluate Each Term of the Series

Calculate each term of the sum separately: - For \( k=1 \), \( \frac{1}{k} = \frac{1}{1} = 1 \).- For \( k=2 \), \( \frac{1}{k} = \frac{1}{2} \).- For \( k=3 \), \( \frac{1}{k} = \frac{1}{3} \).

Step 3: Add the Evaluated Terms

Now add the calculated terms together: \[ 1 + \frac{1}{2} + \frac{1}{3} \].Convert to a common denominator (6) and add:\( \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6} \).

Key Concepts

Arithmetic Series

An arithmetic series is a type of mathematical series in which each term increases by a constant amount from the previous term. However, in the context of the given exercise, we are dealing with a series of fractions, not an arithmetic progression where each term adds a common difference. Nonetheless, in any series where terms are summed, the principles behind arithmetic series can help in understanding the idea of accumulating consecutive terms.

In an arithmetic series, you'd typically see something like \( a + (a+d) + (a+2d) + \ldots + [a+(n-1)d] \), where \( a \) is the first term, \( d \) is the common difference, and \( n \) is the number of terms. Although the sum \( \sum_{k=1}^{3} \frac{1}{k} \) isn't an arithmetic series, recognizing that you are summing over a sequence of terms prepares you for approaching a range of series, whether arithmetic or not.

To sum any arithmetic series quickly, we use the formula:

• Total Sum = \( \frac{n}{2} \times (2a + (n-1)d) \)
• Where \( n \) is total number of terms.

But remember, in series like the one provided, where fractions with different denominators are present, it steps outside the arithmetic series framework but engages similar summation processes.

Fraction Addition

Adding fractions involves combining fractions that may have different denominators. This can sometimes make the process seem complex, but let's break it down. In the exercise, you are tasked to add the fractions \(1\), \(\frac{1}{2}\), and \(\frac{1}{3}\). This means you need to find a way to add their parts systematically to get the total sum.

When adding fractions, you proceed as follows:

• First, check if the fractions have the same denominator.
• If not, convert them to equivalent fractions with the same denominator.
• Add the numerators of these new fractions.

For example, to add \( 1 + \frac{1}{2} + \frac{1}{3} \), you first convert \( 1\) into a fraction with a common denominator. The fractions \(\frac{1}{2}\) and \(\frac{1}{3}\) already have simple forms, but to add them, you will need them to share a common denominator, which leads us to our next concept.

Common Denominators

A crucial step in adding fractions is finding a common denominator, which is essential for "lining up" the fractions properly for addition. When fractions have different denominators, as in this exercise, comparing their values directly is not straightforward.

To solve the problem \( 1 + \frac{1}{2} + \frac{1}{3} \), you convert \( 1 \) into a fraction, \( \frac{6}{6} \) in this case, since the Least Common Multiple (LCM) of 1, 2, and 3 is 6. And then:

• Multiply the numerator and denominator of \( \frac{1}{2} \) by 3 to get \( \frac{3}{6} \).
• Multiply the numerator and denominator of \( \frac{1}{3} \) by 2 to get \( \frac{2}{6} \).

Now, with the common denominator in place, add the numerators: \[ \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6} \]

This gives a single fraction that is the final result of the original sum. By converting to common denominators, you simplify the addition process, making it easy to see the total value.