To prove that A > 2023/2, we can compare A with the integral of the function f(x) = 1/x.

Let's consider the integral of f(x) from 1 to 2024:

∫(1/x) dx from 1 to 2024

= ln(x) from 1 to 2024

= ln(2024) - ln(1)

= ln(2024)

Since ln(x) is a monotonically increasing function, we can conclude that the sum A is greater than the integral of f(x) from 1 to 2024.

Therefore, A > ln(2024).

To compare ln(2024) with 2023/2, we can use the fact that ln(x) < x for x > 0.

ln(2024) < 2024

2024 < 2023/2 * 2

2024 < 4046

Therefore, A > ln(2024) > 2023/2.

Hence, A > 2023/2.