1. **Problem Statement:** Find the sum $$S=\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+\cdots+\frac{1}{n(n+1)}$$ and understand why the term $$\frac{1}{k(k+1)}$$ is split into partial fractions of the form $$\frac{a}{k} - \frac{b}{k+1}$$. 2. **Formula and Concept:** The method used here is called **partial fraction decomposition**. It helps simplify complex rational expressions into simpler fractions that are easier to sum. 3. **Why split into partial fractions?** We want to express: $$\frac{1}{k(k+1)} = \frac{a}{k} - \frac{b}{k+1}$$ for some constants $$a$$ and $$b$$. This is because when we sum these terms from $$k=1$$ to $$n$$, many terms cancel out (telescoping series), making the sum easier to find. 4. **Find constants $$a$$ and $$b$$:** Multiply both sides by $$k(k+1)$$: $$1 = a(k+1) - b k$$ Simplify: $$1 = a k + a - b k = (a - b)k + a$$ Since this must hold for all $$k$$, the coefficient of $$k$$ must be zero and the constant term must be 1: $$a - b = 0 \implies a = b$$ $$a = 1$$ So, $$a = 1$$ and $$b = 1$$. 5. **Rewrite the term:** $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}$$ 6. **Sum the series:** $$S = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$ Notice the telescoping effect: all intermediate terms cancel out, leaving: $$S = 1 - \frac{1}{n+1} = \frac{n}{n+1}$$ **Final answer:** $$S = \frac{n}{n+1}$$ This explains why splitting into $$\frac{a}{k} - \frac{b}{k+1}$$ helps: it creates a telescoping series that simplifies the sum greatly.