Problem 66 - Precalculus

Question

In calculus, partial fractions are used to calculate the sums of infinite series. Find the partial-fraction decomposition of the summand. $$\sum_{k=1}^{\infty} \frac{1}{k(k+1)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{k} - \frac{1}{k+1} \). The series sum is 1.

Step 1: Understand the Problem

We need to find the partial fraction decomposition of the expression \( \frac{1}{k(k+1)} \) associated with the series sum \( \sum_{k=1}^{\infty} \frac{1}{k(k+1)} \). Partial fraction decomposition expresses a rational function as a sum of simpler fractions.

Step 2: Set Up the Partial Fraction Decomposition

The expression \( \frac{1}{k(k+1)} \) needs to be decomposed into simpler fractions. Assume:\[\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}\]where \( A \) and \( B \) are constants that we need to determine.

Step 3: Solve for Constants A and B

Multiply both sides by \( k(k+1) \) to eliminate the denominators:\[1 = A(k+1) + Bk\]Expand and rearrange the terms:\[1 = Ak + A + Bk\]Combine like terms:\[1 = (A + B)k + A\]For the equation to be true for all \( k \), equate the constant terms and the coefficients of \( k \):1. \( A + B = 0 \)2. \( A = 1 \)Solve these equations to find: \( A = 1 \), \( B = -1 \).

Step 4: Rewrite the Expression

Substitute \( A = 1 \) and \( B = -1 \) back into the partial fraction decomposition:\[\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}\]

Step 5: Observe the Telescoping Series Effect

Upon inserting the partial fractions into the series:\[\sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{k+1} \right)\]Notice that this is a telescoping series where terms cancel out successively. For each pair, \( \frac{1}{k+1} \) from one term cancels with \( \frac{1}{k+1} \) from the next term.

Step 6: Conclude the Series Sum

As the telescoping series progresses, nearly all terms cancel, except the first term of the series. Thus, as \( n \to \infty \), the sum tends towards:\[\lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1\]

Key Concepts

Telescoping Series

Telescoping Series are a magical way of simplifying infinite series. They are like magical chains where each link cancels out the next, leaving only a few bits at the end. Let's take a closer look. When a series telescopes, it means that many terms end up cancelling each other out, making it super easy to find the sum. 🔍 For example, look at our partial fractions: \( \frac{1}{k} - \frac{1}{k+1} \). Here’s what’s happening:

• When you start adding these fractions from \( k=1 \) to infinity, each \( \frac{1}{k+1} \) cancels out the \( \frac{1}{k+1} \) from the previous term.
• This process of cancellation simplifies the series greatly.
• Ultimately, you’re left with just the first and last pieces, which usually makes finding the series sum much more transparent.

This type of series is super useful because you don’t need to keep track of a zillion terms. You focus only on the ones that aren’t swallowed up by the cancelling game! ⚖️

Infinite Series

Infinite series might seem a bit tricky at first, but they're all about adding up numbers that keep going forever. Although it seems impossible to get a sum for infinite numbers, mathematicians have found brilliant ways to deal with them.
The idea is to see if the sum gets closer and closer to a particular number as you add more terms. This concept is called convergence. If an infinite series "converges," it means that by adding an increasing number of terms, we get nearer to a specific number.

• In some cases, like our telescoping series, the terms simplify so much that you can see where they will land as more and more terms are added.
• The sum, as you might guess, relies largely on how the terms of a series behave as they grow larger.

In contrast, if the series keeps growing wildly without settling down, it diverges, meaning it doesn't have a finite sum. Learning to identify a converging series, especially through telescoping, is a rewarding mathematical journey where you find powerful shortcuts.

Rational Functions

Rational functions, like old friends, pop up frequently in math problems, especially when dealing with series or calculus. Essentially, a rational function is a fraction where both the numerator and the denominator are polynomials.
These functions are the foundation of partial fraction decomposition. For any given rational function, breaking it into simpler fractions makes it easier to work with, especially for integration or summing series.

• For our example \( \frac{1}{k(k+1)} \), we decompose it into simpler terms \( \frac{A}{k} + \frac{B}{k+1} \).
• Solving these equations means finding values for \( A \) and \( B \) that make this equality true for all values of \( k \).

This approach turns complex problems into a series of easily manageable steps, making calculations more straightforward and less intimidating. Mastering rational functions and their decomposition is like being handed a powerful tool to find simplicity and order amidst complexity.