1. **Problem Statement:** Prove by mathematical induction that for all positive integers $n$, $$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}.$$ 2. **Base Case ($n=1$):** Evaluate the left-hand side (LHS): $$\sum_{r=1}^1 \frac{1}{r(r+1)} = \frac{1}{1 \cdot 2} = \frac{1}{2}.$$ Evaluate the right-hand side (RHS): $$\frac{1}{1+1} = \frac{1}{2}.$$ Since LHS = RHS, the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some positive integer $k$, that is, $$\sum_{r=1}^k \frac{1}{r(r+1)} = \frac{k}{k+1}.$$ 4. **Inductive Step:** We need to prove the formula holds for $k+1$, i.e., $$\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \frac{k+1}{k+2}.$$ Start with the left side: $$\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \left(\sum_{r=1}^k \frac{1}{r(r+1)}\right) + \frac{1}{(k+1)(k+2)}.$$ By the inductive hypothesis, this equals: $$\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}.$$ Combine the terms over the common denominator $(k+1)(k+2)$: $$\frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)}.$$ Simplify the numerator: $$k(k+2) + 1 = k^2 + 2k + 1 = (k+1)^2.$$ So the expression becomes: $$\frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}.$$ This matches the right side for $k+1$. 5. **Conclusion:** By mathematical induction, the formula $$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}$$ holds for all positive integers $n$.