Problem 4 - The Calculus with Analytic Geometry

Question

\(\sum_{j=3}^{6} \frac{2}{j(j-2)}\)

Step-by-Step Solution

Answer

\frac{91}{30}

Step 1: Understand the Summation

The given summation is \(\begin{equation*} \sum_{j=3}^{6} \frac{2}{j(j-2)} \end{equation*}\). This notation means that you need to substitute the values of \( j \) from 3 to 6 into the expression \( \frac{2}{j(j-2)} \) and then sum the results.

Step 2: Simplify the Fraction

Rewrite \( \frac{2}{j(j-2)} \) in a simpler form by partial fraction decomposition. Suppose: \(\begin{equation*} \frac{2}{j(j-2)} = \frac{A}{j} + \frac{B}{j-2} \end{equation*}\). Solve for A and B by setting up the equation: \(\begin{equation*} 2 = A(j-2) + B(j) \end{equation*}\). By equating coefficients, you find that A = 1 and B = 1, thus \(\begin{equation*} \frac{2}{j(j-2)} = \frac{1}{j} + \frac{1}{j-2} \end{equation*}\).

Step 3: Substitute and Expand the Summation

Use the partial fractions in the summation: \(\begin{equation*} \sum_{j=3}^{6} \frac{2}{j(j-2)} = \sum_{j=3}^{6} \bigg( \frac{1}{j} + \frac{1}{j-2} \bigg) \end{equation*}\). This split into two separate summations: \(\begin{equation*} \sum_{j=3}^{6} \frac{1}{j} + \sum_{j=3}^{6} \frac{1}{j-2} \end{equation*}\).

Step 4: Evaluate Each Summation

Evaluate each part: \(\begin{equation*} \sum_{j=3}^{6} \frac{1}{j} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \end{equation*}\) and \(\begin{equation*} \sum_{j=3}^{6} \frac{1}{j-2} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \end{equation*}\).

Step 5: Sum the Results

Add the terms: \(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\) results in \(\frac{20}{60} + \frac{15}{60} + \frac{12}{60} + \frac{10}{60} = \frac{57}{60}\). Similarly, \(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\) becomes \(\frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12} = \frac{25}{12}\).

Step 6: Find the Total Summation

Combine the two sums: \(\frac{57}{60} + \frac{25}{12}\). Convert to a common denominator: \(\frac{57}{60} = \frac{57}{60} = \frac{57}{60}\) and \(\frac{25}{12} = \frac{125}{60} \). Adding these gives \(\frac{57}{60} + \frac{125}{60} = \frac{182}{60} = \frac{91}{30}\). This fraction cannot be simplified further, so the final answer is \(\frac{91}{30}\).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler ones. This is particularly useful in calculus for simplifying the summation and integration of functions. Take the expression \(\frac{2}{j(j-2)}\).
First, we need to assume that this can be written as \( \frac{A}{j} + \frac{B}{j-2} \). To find A and B, multiply through by the denominator \( j(j-2) \), giving us \(2 = A(j-2) + Bj\).
Next, by comparing coefficients, we can solve for A and B. In our case, this results in \( A = 1 \) and \( B = 1 \), so \( \frac{2}{j(j-2)} = \frac{1}{j} + \frac{1}{j-2} \).
This decomposition simplifies our original sum and makes it easier to handle.

Evaluating Summations

Summations are a way to add up a sequence of numbers based on a formula. For example, the summation \(\sum_{j=3}^{6} \frac{2}{j(j-2)} \) means that we sum the expression \( \frac{2}{j(j-2)} \) as j takes on values from 3 to 6. Using partial fraction decomposition, we simplify this to \(\frac{1}{j} + \frac{1}{j-2} \).
This divides our summation into two simpler series: \( \sum_{j=3}^{6} \frac{1}{j} + \sum_{j=3}^{6} \frac{1}{j-2} \).
We then substitute each value of j into both summations and sum the results:

• For \(\sum_{j=3}^{6} \frac{1}{j} \), we get \(\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\).
• For \(\sum_{j=3}^{6} \frac{1}{j-2} \), we get \(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\).

Finally, adding these two sums together gives us the value of the original summation.

Series Expansion

Series expansion allows us to represent functions as sums of simpler terms. This can be extremely helpful when dealing with complicated expressions. In our original problem, we expanded \(\frac{2}{j(j-2)} \) into simpler fractions \( \frac{1}{j} + \frac{1}{j-2} \).
By doing this, we transformed a difficult problem into smaller, more manageable pieces.
This technique also aids in recognizing potential telescoping series, where intermediate terms cancel out, leaving a much simpler sum. While our example didn’t completely telescope, breaking it into parts still significantly simplified what we had to compute manually.
Using partial fraction decomposition and understanding how to handle summations and series is essential for tackling a wide range of calculus problems effectively.