To prove the inequality \( \frac{11}{15} < A < \frac{3}{2} \) where \( A = \sum_{n=21}^{60} \frac{1}{n} \), we can use the properties of the harmonic series and integrals to approximate 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 for \( A \), we can compare the sum to an integral. The harmonic series can be approximated by the natural logarithm function. Specifically, we can use the integral test for convergence:

\[ \int_{21}^{61} \frac{1}{x} \, dx < \sum_{n=21}^{60} \frac{1}{n} < \int_{20}^{60} \frac{1}{x} \, dx \]

Evaluating these integrals, we get:

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

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

Using the fact that \( \ln(3) \approx 1.0986 \) and \( \ln\left(\frac{61}{21}\right) \approx 1.0986 \), we see that:

\[ \sum_{n=21}^{60} \frac{1}{n} \approx \ln\left(\frac{61}{21}\right) \approx 1.0986 \]

However, this approximation is slightly less than \( \frac{3}{2} \approx 1.5 \). Therefore, we can conclude that:

\[ A < \frac{3}{2} \]

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

To find a lower bound for \( A \), we can use a similar approach. We know that:

\[ \int_{21}^{61} \frac{1}{x} \, dx < \sum_{n=21}^{60} \frac{1}{n} \]

Evaluating the integral:

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

We need to show that this value is greater than \( \frac{11}{15} \approx 0.7333 \).

Since \( \ln\left(\frac{61}{21}\right) \approx 1.0986 \), which is clearly greater than \( 0.7333 \), we have:

\[ \frac{11}{15} < \sum_{n=21}^{60} \frac{1}{n} \]

Thus, combining both bounds, we have:

\[ \frac{11}{15} < A < \frac{3}{2} \]

Therefore, we have successfully proven the inequality \( \frac{11}{15} < A < \frac{3}{2} \).