To prove the inequality \( \frac{11}{15} < A < \frac{3}{2} \) for \( A = \sum_{n=21}^{60} \frac{1}{n} \), we can use the properties of harmonic series and integrals to estimate the sum.

First, let's rewrite \( A \) as:
\[ A = \sum_{n=21}^{60} \frac{1}{n} \]

### Upper Bound (\( A < \frac{3}{2} \))

To find an upper bound, we can compare the sum to an integral. The harmonic series can be approximated by the natural logarithm function. Specifically, the sum of reciprocals can be bounded by an integral:
\[ \sum_{n=a}^{b} \frac{1}{n} < \int_{a-1}^{b} \frac{1}{x} \, dx \]

For our case:
\[ \sum_{n=21}^{60} \frac{1}{n} < \int_{20}^{60} \frac{1}{x} \, dx \]

Evaluating the integral:
\[ \int_{20}^{60} \frac{1}{x} \, dx = \ln(60) - \ln(20) = \ln\left(\frac{60}{20}\right) = \ln(3) \]

Using the approximation \( \ln(3) \approx 1.0986 \):
\[ \sum_{n=21}^{60} \frac{1}{n} < 1.0986 \]

Since \( 1.0986 < \frac{3}{2} = 1.5 \), we have:
\[ A < \frac{3}{2} \]

### Lower Bound (\( \frac{11}{15} < A \))

To find a lower bound, we can use a similar approach but with a different integral:
\[ \sum_{n=21}^{60} \frac{1}{n} > \int_{21}^{61} \frac{1}{x} \, dx \]

Evaluating the integral:
\[ \int_{21}^{61} \frac{1}{x} \, dx = \ln(61) - \ln(21) = \ln\left(\frac{61}{21}\right) \]

Using the approximation \( \ln\left(\frac{61}{21}\right) \approx 1.0498 \):
\[ \sum_{n=21}^{60} \frac{1}{n} > 1.0498 \]

Now, we need to show that \( 1.0498 > \frac{11}{15} \):
\[ \frac{11}{15} \approx 0.7333 \]

Clearly, \( 1.0498 > 0.7333 \), so:
\[ \frac{11}{15} < A \]

### Conclusion

Combining both bounds, we have:
\[ \frac{11}{15} < A < \frac{3}{2} \]

Thus, we have successfully proven the inequality:
\[ \frac{11}{15} < \sum_{n=21}^{60} \frac{1}{n} < \frac{3}{2} \]