1. **Problem statement:** Prove by mathematical induction that the harmonic series partial sum satisfies $$H_{2n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2n} \geq 1 + \frac{n}{2}$$ for all positive integers $n$. 2. **Base case:** For $n=1$, we have $$H_2 = 1 + \frac{1}{2} = 1.5$$ and $$1 + \frac{1}{2} = 1.5,$$ so $$H_2 \geq 1 + \frac{1}{2}$$ holds. 3. **Inductive hypothesis:** Assume for some $n = k$ that $$H_{2k} \geq 1 + \frac{k}{2}.$$ 4. **Inductive step:** We want to prove $$H_{2(k+1)} \geq 1 + \frac{k+1}{2}.$$ 5. Write $$H_{2(k+1)} = H_{2k} + \frac{1}{2k+1} + \frac{1}{2k+2}.$$ 6. Using the inductive hypothesis, substitute $$H_{2k} \geq 1 + \frac{k}{2}$$ to get $$H_{2(k+1)} \geq 1 + \frac{k}{2} + \frac{1}{2k+1} + \frac{1}{2k+2}.$$ 7. We need to show $$\frac{1}{2k+1} + \frac{1}{2k+2} \geq \frac{1}{2}.$$ 8. Find a common denominator and sum: $$\frac{1}{2k+1} + \frac{1}{2k+2} = \frac{(2k+2) + (2k+1)}{(2k+1)(2k+2)} = \frac{4k+3}{(2k+1)(2k+2)}.$$ 9. We want to prove $$\frac{4k+3}{(2k+1)(2k+2)} \geq \frac{1}{2}.$$ 10. Cross-multiplied: $$2(4k+3) \geq (2k+1)(2k+2).$$ 11. Expand both sides: $$8k + 6 \geq 4k^2 + 6k + 2.$$ 12. Rearrange: $$0 \geq 4k^2 - 2k - 4,$$ which is $$4k^2 - 2k - 4 \leq 0.$$ 13. Check if this inequality holds for all $k \geq 1$: Calculate the quadratic at $k=1$: $$4(1)^2 - 2(1) - 4 = 4 - 2 - 4 = -2 \leq 0,$$ which is true. 14. Since the quadratic opens upward (coefficient of $k^2$ is positive), and it is negative at $k=1$, the inequality holds for $k=1$ but fails for larger $k$. 15. Therefore, the inequality $$\frac{1}{2k+1} + \frac{1}{2k+2} \geq \frac{1}{2}$$ does not hold for all $k$, so the original statement is false. 16. **Conclusion:** The inequality $$H_{2n} \geq 1 + \frac{n}{2}$$ is not true for all $n$. The base case holds, but the inductive step fails. **Final answer:** The statement $$H_{2n} \geq 1 + \frac{n}{2}$$ is false and cannot be proven by induction.