1. **State the problem:** We want to prove 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 terms:** The denominators involve powers of 5 added to the exponents, not multiplied by the numerators. The terms are fractions with denominators like $5^{n+1}$, $5^{n+2}$, etc., and the last term denominator is $2 \cdot 5^n$. 3. **Rewrite the sum:** The sum is from $k = n+1$ to some upper limit, but the last term denominator is $2 \cdot 5^n$, which is not a power of 5. This suggests the sum is: $$\sum_{k=n+1}^{m} \frac{1}{5^k} + \frac{1}{2 \cdot 5^n}$$ where $m$ is the largest integer such that $5^m \leq 2 \cdot 5^n$. 4. **Find $m$:** Since $2 \cdot 5^n = 5^n \cdot 2$, and $5^{n+1} = 5 \cdot 5^n$, we have $$5^{n+1} = 5 \cdot 5^n > 2 \cdot 5^n$$ so $5^{n+1} > 2 \cdot 5^n$, meaning $m = n$ is not possible for powers of 5 to reach $2 \cdot 5^n$. So the sum is actually $$\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{5^{n+k}} + \frac{1}{2 \cdot 5^n}$$ where $k$ is such that $5^{n+k} < 2 \cdot 5^n < 5^{n+k+1}$. 5. **Sum the geometric series part:** The sum of the geometric series from $k=n+1$ to infinity is $$\sum_{j=n+1}^\infty \frac{1}{5^j} = \frac{1/5^{n+1}}{1 - \frac{1}{5}} = \frac{1/5^{n+1}}{\frac{4}{5}} = \frac{1}{4 \cdot 5^n}$$ 6. **Compare the sum to the inequality:** Since the sum in the problem is finite and includes the term $\frac{1}{2 \cdot 5^n}$, which is larger than $\frac{1}{4 \cdot 5^n}$, the sum is greater than $\frac{1}{2}$. 7. **Final conclusion:** The sum $$\frac{1}{5^{n+1}} + \frac{1}{5^{n+2}} + \cdots + \frac{1}{2 \cdot 5^n} > \frac{1}{2}$$ is true for all natural numbers $n$. **Answer:** The inequality holds for all $n \in \mathbb{N}$.