The problem defines a sequence $a_n$ with the formula:

$a_n = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)}$

This type of series is telescoping, where each term can be simplified to reveal a pattern that makes summing easier.

To evaluate $a_n$, let’s break down each term in the series:

$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$

Using partial fraction decomposition, each term in the series becomes:

$\frac{1}{1 \cdot 2} = \frac{1}{1} - \frac{1}{2}, \quad \frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}, \quad \text{and so on,} \quad \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$

So, the sequence $a_n$ becomes:

$a_n = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right)$

In this telescoping series, most terms cancel out, leaving only:

$a_n = 1 - \frac{1}{n+1}$

Therefore,

$a_n = \frac{n}{n+1}$

Would you like further details on any step or have any questions?

Here are five related questions that build on this solution:

1. How would you find the limit of $a_n$ as $n \to \infty$?
2. Can you express $a_n$ as a continuous function of $n$?
3. How would the series change if the general term was $\frac{1}{n(n+2)}$?
4. What happens if the sum goes to infinity? Would it converge or diverge?
5. How could you verify the solution by induction?

Tip: Telescoping series often simplify complex sums by breaking terms down into parts that cancel sequentially.