1. **Problem statement:** Prove by mathematical induction that $$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$ for all positive integers $n$. 2. **Base Case ($n=1$):** Calculate the left side: $$\sum_{i=1}^1 \frac{1}{i(i+1)} = \frac{1}{1\times 2} = \frac{1}{2}$$ Calculate the right side: $$\frac{1}{1+1} = \frac{1}{2}$$ Both sides equal $\frac{1}{2}$, so the base case holds. 3. **Inductive Hypothesis:** Assume the formula holds for some $k \geq 1$: $$\sum_{i=1}^k \frac{1}{i(i+1)} = \frac{k}{k+1}$$ 4. **Inductive Step:** Prove it holds for $k+1$: Consider: $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \left(\sum_{i=1}^k \frac{1}{i(i+1)}\right) + \frac{1}{(k+1)(k+2)}$$ Using the inductive hypothesis: $$= \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$ Find a common denominator $ (k+1)(k+2)$: $$= \frac{k(k+2)}{(k+1)(k+2)} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)}$$ Simplify the numerator: $$k(k+2) + 1 = k^2 + 2k + 1 = (k+1)^2$$ So: $$= \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}$$ This matches the formula for $n = k + 1$. 5. **Conclusion:** By the principle of mathematical induction, the statement is true for all positive integers $n$: $$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$$