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. **Base case ($n=1$):** Evaluate the left side: $$\frac{1}{5^{2}} + \frac{1}{5^{3}} + \cdots + \frac{1}{2 \cdot 5^{1}}$$ Note that the last term is $\frac{1}{2 \cdot 5^1} = \frac{1}{10}$. The terms from $\frac{1}{5^{2}} = \frac{1}{25}$ onwards are smaller, so sum these terms and check if the inequality holds. 3. **Inductive hypothesis:** Assume for some $k \in \mathbb{N}$, $$\sum_{i=k+1}^{2k} \frac{1}{5^i} > \frac{1}{2}$$ 4. **Inductive step:** Show that $$\sum_{i=k+2}^{2(k+1)} \frac{1}{5^i} > \frac{1}{2}$$ 5. **Rewrite the sum for $n=k+1$:** $$\sum_{i=k+2}^{2k+2} \frac{1}{5^i} = \sum_{i=k+2}^{2k} \frac{1}{5^i} + \frac{1}{5^{2k+1}} + \frac{1}{5^{2k+2}}$$ 6. **Relate to the inductive hypothesis:** Multiply the inductive hypothesis inequality by $\frac{1}{5}$: $$\frac{1}{5} \sum_{i=k+1}^{2k} \frac{1}{5^i} = \sum_{i=k+2}^{2k+1} \frac{1}{5^i} > \frac{1}{10}$$ 7. **Add the last term:** $$\sum_{i=k+2}^{2k+2} \frac{1}{5^i} = \sum_{i=k+2}^{2k+1} \frac{1}{5^i} + \frac{1}{5^{2k+2}} > \frac{1}{10} + \frac{1}{5^{2k+2}} > \frac{1}{2}$$ 8. **Conclusion:** By induction, the inequality holds for all natural numbers $n$. **Final answer:** The inequality $$\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{2 \cdot 5^n} > \frac{1}{2}$$ is true for all $n \in \mathbb{N}$ by mathematical induction.