Question

Find all the clements of (0,1,2,3\\} that, when substituted for \(n,\) satisfy: $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}$$

Short answer

All elements \(n \in \{0, 1, 2, 3\}\) satisfy the equation.

Step-by-step solution

Understand the Equation

The equation given is a sum of fractions: \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}\). This can be simplified by noticing that \(\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\), which is a telescoping series.

Rewrite the Series Using Telescoping Property

The series can be rewritten as \(\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\). This simplifies to \(1 - \frac{1}{n+1}\) because all intermediate terms cancel out.

Simplify Right Hand Side of the Equation

The right hand side of the equation is \(\frac{n}{n+1}\). This can also be rewritten as \(1 - \frac{1}{n+1}\). Thus, the simplified form is exactly the same as what we obtained from the telescoping series.

Match Simplified Forms to Identify Valid n

Both sides of the equation simplify to \(1 - \frac{1}{n+1}\). Therefore, \(n\) can be any value where this equivalence holds true.

Conclusion for Possible Values of n

Given the simplification, the original equation is always true for any integer \(n\) because each side simplifies to the same expression. Checking for values in the set \(\{0, 1, 2, 3\}\), we see any integer \(n\) would result in the same equivalences.

Key concepts

Discrete Mathematics

Discrete Mathematics is a branch of mathematics that deals with distinct and separate objects. It includes topics like set theory, graph theory, and logic. Unlike continuous mathematics, which studies smoothly varying phenomena, discrete mathematics focuses on items that can often be counted using integers.

Telescoping series, like the one in our problem, are frequently found in discrete mathematics. These come into play when you need to find patterns in sequences or solve puzzles requiring combinatorial methods.

When handling discrete objects, properties such as integer partitions and relations between sets become vital in understanding the structure and behavior of these numbers. The telescoping sequence is a great example of how discrete mathematics helps simplify complex summation problems by systematically canceling out terms.

Summation

Summation is the process of adding a sequence of numbers. It is a critical concept in both algebra and calculus. However, in discrete mathematics, summation often involves series that are not continuous.

In the given problem, a sum of fractions is involved, known formally as a telescoping series. In a telescoping series, many terms cancel each other out, making the summation much easier to handle. You start with a complex sum and break it down, cancelling intermediate terms until you are left with only the first and last terms.

This exercise demonstrates how the sum of the series \(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)}\) simplifies to a much simpler expression \(1 - \frac{1}{n+1}\), emphasizing the elegance of telescoping series.

• Step 1: Recognize the pattern and express each term as \(\frac{1}{k} - \frac{1}{k+1}\).
• Step 2: Observe how each subtractive element cancels with a part of the subsequent term.
• Step 3: Simplify to find the remaining terms after the cancellation.

This approach shows how understanding series summation is crucial for solving complex mathematical problems efficiently.

Fraction Simplification

Fraction Simplification is the process of reducing fractions to their simplest form. In this exercise, simplifying fractions plays a crucial role in resolving the equation. By recognizing that \(\frac{1}{k(k+1)}\) can be rewritten as \(\frac{1}{k} - \frac{1}{k+1}\), it transforms the series into one that is easier to handle.

This clever rearrangement is a key feature of telescoping series and significantly aids in simplifying expressions.

Here’s how fraction simplification works in this context:

• Identify the product in the denominator, \(k(k+1)\).
• Rewrite it as a difference of fractions using partial fraction decomposition, \(\frac{1}{k} - \frac{1}{k+1}\).
• Explore how the new form allows intermediate terms to cancel each other, leading to a simplified final expression.

While fraction simplification might seem daunting at first, especially with complex denominators, it often leads to clarity and understanding in mathematical expressions.