1. **State the problem:** We want to prove by induction that for all positive integers $n$, $$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}.$$ 2. **Base case:** When $n=1$, the left side is: $$\sum_{r=1}^1 \frac{1}{r(r+1)} = \frac{1}{1 \cdot 2} = \frac{1}{2}.$$ The right side is: $$\frac{1}{1+1} = \frac{1}{2}.$$ Since both sides equal $\frac{1}{2}$, the base case holds. 3. **Induction hypothesis:** Assume that the formula is true for some positive integer $k$, i.e., $$\sum_{r=1}^k \frac{1}{r(r+1)} = \frac{k}{k+1}.$$ This is our induction assumption. 4. **Inductive step:** We need to show that the formula is true for $k+1$, i.e., $$\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \frac{k+1}{k+2}.$$ Starting from 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 induction hypothesis, substitute the sum up to $k$: $$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}.$$ 5. **Simplify the right side:** Find a common denominator $ (k+1)(k+2)$: $$= \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)}.$$ Calculate the numerator: $$k(k+2)+1 = k^2 + 2k + 1 = (k+1)^2.$$ So the expression becomes: $$\frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}.$$ 6. **Conclusion:** We have shown: $$\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \frac{k+1}{k+2},$$ which completes the induction step. Therefore, by mathematical induction, the formula $$\sum_{r=1}^n \frac{1}{r(r+1)} = \frac{n}{n+1}$$ holds for all positive integers $n$. This proof relies on the idea that each term can be decomposed and telescoped in the sum, simplifying the overall expression nicely.