Question

Find a nice formula for the sum $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)} $$

Short answer

The sum is \( 1 - \frac{1}{n+1} \).

Step-by-step solution

Identify the Pattern

Observe the pattern in the given series: \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} \). Each term of the series is of the form \( \frac{1}{k(k+1)} \) for \( k = 1 \) to \( n \).

Simplify a General Term

Recognize that \( \frac{1}{k(k+1)} \) can be simplified using partial fractions. We can express it as: \( \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \).

Write the Expanded Series

Substitute the partial fraction decomposition into the series: \( \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \).

Observe the Telescoping Effect

Notice that the series has a telescoping nature, where most terms will cancel out: \( 1 - \frac{1}{n+1} \).

Write the Simplified Sum

After cancelation, the series simplifies to: \( 1 - \frac{1}{n+1} \). Thus, the sum of the series \( \sum_{k=1}^n \frac{1}{k(k+1)} \) is equal to this expression.

Key concepts

Partial Fraction Decomposition

When faced with complex expressions, partial fraction decomposition becomes a valuable tool. In this exercise, we use it to simplify a term of the series: \( \frac{1}{k(k+1)} \). The idea here is to break down this fraction into simpler parts that are easier to manage.
Partial fraction decomposition allows us to rewrite \( \frac{1}{k(k+1)} \) into the form \( \frac{1}{k} - \frac{1}{k+1} \). This transformation is very useful because it sets the stage for simplifying the whole series later on.
Let's consider what happens:

• We identify \( \frac{1}{k(k+1)} \) and determine that it can be rewritten as two separate fractions.
• This deceptively simple move from one fraction into a subtraction of two fractions unravels the complexity, making our task of summing easier.

Ultimately, partial fraction decomposition is an essential step in this process because it unveils the relationships between each term in the sequence.

Series Summation

Once we have simplified each term using partial fractions, the series summation begins. We focus on adding up the sequence of simplified terms:\
\[\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right)\]
This series involves a range from \( k=1 \) to \( k=n \), with each term crafted from our decomposition.
To understand the series summation, it helps to:

• Recognize that the series now is a sequence of pairs of subtractions.
• Each subtraction \( \frac{1}{k} - \frac{1}{k+1} \) plays an integral part in simplifying the overall sum.

The goal of summing everything up is to consolidate these interactions into a single, simple expression that represents the total of our series.

Telescoping Nature

The telescoping nature of a series is like magic unfolding. It describes how, in a cleverly structured series, most terms cancel each other out. This property significantly simplifies the task of summing a sequence.
In our example, this is visible as you expand the series into:
\[\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right)\]
Here's what happens:

• Each \( -\frac{1}{k+1} \) term cancels out with the \( \frac{1}{k+1} \) of the following pair.
• By canceling terms, you greatly reduce complexity, leaving behind only the first term of the series and one part of the last term.

This leaves us with \( 1 - \frac{1}{n+1} \), a beautifully simple result that neatly ties up the series.
The telescoping effect is a powerful mathematical property that makes evaluating specific types of series much more manageable.