If \(\displaystyle \left( \frac{1}{\alpha+1} + \frac{1}{\alpha+2} + \cdots + \frac{1}{\alpha+1012} \right) - \left( \frac{1}{2\cdot 1} + \frac{1}{4\cdot 3} + \frac{1}{6\cdot 5} + \cdots + \frac{1}{2024\cdot 2023} \right) = \frac{1}{2024} \), then \( \alpha \) is equal to _________.
