Question: 1 \over n + 1 + 1 \over n + 2 + \cdots + 1 \over 2n > 13 \over 24
1. **Stating the problem:** We want to prove and explain why the sum of fractions $$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}$$ is greater than $$\frac{13}{24}$$ for some positive integer $n$.
2. **Understanding the sum:** The sum is a partial sum of the harmonic series, specifically the terms from $\frac{1}{n+1}$ up to $\frac{1}{2n}$. We want to show:
$$\sum_{k=n+1}^{2n} \frac{1}{k} > \frac{13}{24}$$
3. **Key idea - Integral test and comparison:** The harmonic series grows slowly but diverges. The sum from $n+1$ to $2n$ can be compared to an integral:
$$\int_{n+1}^{2n+1} \frac{1}{x} dx = \ln(2n+1) - \ln(n+1) = \ln\left(\frac{2n+1}{n+1}\right)$$
Since $\frac{2n+1}{n+1} > 2$, this integral is greater than $\ln(2) \approx 0.693$.
4. **Lower bound for the sum:** The sum is greater than the integral from $n+1$ to $2n+1$ minus the first term:
$$\sum_{k=n+1}^{2n} \frac{1}{k} > \int_{n+1}^{2n+1} \frac{1}{x} dx - \frac{1}{n+1} = \ln\left(\frac{2n+1}{n+1}\right) - \frac{1}{n+1}$$
For $n \geq 1$, this is still greater than some positive constant.
5. **Checking for $n=1$:**
$$\frac{1}{2} = 0.5$$ which is less than $\frac{13}{24} \approx 0.5417$.
6. **Checking for $n=2$:**
$$\frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} = 0.5833 > 0.5417$$
So for $n=2$, the inequality holds.
7. **Conclusion:** For all $n \geq 2$, the sum $$\sum_{k=n+1}^{2n} \frac{1}{k} > \frac{13}{24}$$ holds true.
**Final answer:** The inequality is true for $n \geq 2$.