1. **State the problem:** Prove by induction that for all natural numbers $n$, the inequality $$\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{2 \cdot 5^n} > \frac{1}{2}$$ holds. 2. **Clarify the sum:** The sum is from $k = n+1$ to some upper limit. Since the last term is $\frac{1}{2 \cdot 5^n}$, note that $\frac{1}{2 \cdot 5^n} = \frac{1}{2} \cdot \frac{1}{5^n}$, which is larger than $\frac{1}{5^{n+1}}$ because $5^{n+1} = 5 \cdot 5^n$ and $\frac{1}{2} > \frac{1}{5}$. This suggests the sum includes terms from $\frac{1}{5^{n+1}}$ up to $\frac{1}{2 \cdot 5^n}$, but since $\frac{1}{2 \cdot 5^n}$ is not a power of 5 denominator, we interpret the sum as geometric series terms plus the last term. 3. **Rewrite the sum:** The sum is $$S = \sum_{k=n+1}^{\infty} \frac{1}{5^k}$$ truncated or adjusted to include $\frac{1}{2 \cdot 5^n}$ as the last term. But since the problem states the sum ends at $\frac{1}{2 \cdot 5^n}$, we consider the sum $$S = \frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{2 \cdot 5^n}$$ which is a finite sum plus one extra term. 4. **Base case ($n=1$):** Calculate the sum $$S = \frac{1}{5^{2}} + \frac{1}{5^{3}} + \cdots + \frac{1}{2 \cdot 5^{1}} = \frac{1}{25} + \frac{1}{125} + \cdots + \frac{1}{10}$$ Since $\frac{1}{10} = \frac{1}{2 \cdot 5^1}$ is the last term. 5. **Sum of geometric series:** The terms $\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots$ form a geometric series with first term $a = \frac{1}{5^{n+1}}$ and ratio $r = \frac{1}{5}$. Sum of infinite geometric series: $$S_\infty = \frac{a}{1-r} = \frac{\frac{1}{5^{n+1}}}{1 - \frac{1}{5}} = \frac{\frac{1}{5^{n+1}}}{\frac{4}{5}} = \frac{1}{5^{n+1}} \cdot \frac{5}{4} = \frac{1}{4 \cdot 5^n}$$ 6. **Compare sum to $\frac{1}{2}$:** Note that $$\frac{1}{2} > \frac{1}{4 \cdot 5^n}$$ for all $n \geq 1$, so the infinite sum is less than $\frac{1}{2}$, contradicting the inequality. 7. **Re-examine problem statement:** The problem likely means the sum $$\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{2 \cdot 5^n}$$ where the last term is $\frac{1}{2 \cdot 5^n}$, which is larger than $\frac{1}{5^{n+1}}$, so the sum is finite and includes terms up to $\frac{1}{2 \cdot 5^n}$. 8. **Induction proof:** **Base case ($n=1$):** $$S = \frac{1}{5^{2}} + \frac{1}{2 \cdot 5^{1}} = \frac{1}{25} + \frac{1}{10} = \frac{2}{50} + \frac{5}{50} = \frac{7}{50} = 0.14 > 0.5?$$ No, so base case fails if interpreted this way. 9. **Conclusion:** The problem as stated is ambiguous or incorrect as the sum terms and inequality do not hold for base case. **If the problem is to prove by induction that** $$\sum_{k=n+1}^{2n} \frac{1}{5^k} > \frac{1}{2}$$ then proceed as follows: 10. **Assume the problem is:** $$\sum_{k=n+1}^{2n} \frac{1}{5^k} > \frac{1}{2}$$ **Base case ($n=1$):** $$\sum_{k=2}^{2} \frac{1}{5^k} = \frac{1}{25} = 0.04 < 0.5$$ No, so no. 11. **Alternative interpretation:** The problem likely has a typo or needs clarification. **Final:** Please clarify the problem statement for accurate induction proof.