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

 $$\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$.

$$S_2 = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{2023} - \frac{1}{2024} \right)$$

This is the standard expansion for the alternating harmonic series up to $$2024$$:

$$S_2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{2023} - \frac{1}{2024}$$

$$1 - \frac{1}{2} + \frac{1}{3} - \dots - \frac{1}{2n} = \left( 1 + \frac{1}{2} + \dots + \frac{1}{2n} \right) - 2 \left( \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2n} \right)$$

$$= \left( 1 + \frac{1}{2} + \dots + \frac{1}{2n} \right) - \left( 1 + \frac{1}{2} + \dots + \frac{1}{n} \right)$$

$$= \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n}$$

For our case, $$2n = 2024$$, so $$n = 1012$$:

$$S_2 = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2024}$$

$$\left( \sum_{k=1}^{1012} \frac{1}{\alpha+k} \right) - \left( \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + \frac{1}{2024} \right) = \frac{1}{2024}$$

Adding $$\frac{1}{2024}$$ to the $$S_2$$ sum on the right:

$$\sum_{k=1}^{1012} \frac{1}{\alpha+k} = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + \frac{1}{2024} + \frac{1}{2024}$$

Actually, looking at the term $$\frac{1}{2024}$$ on the RHS, if we move the $$S_2$$ terms over:

$$\sum_{k=1}^{1012} \frac{1}{\alpha+k} = \frac{1}{1013} + \frac{1}{1014} + \dots + \frac{1}{2023} + \left( \frac{1}{2024} + \frac{1}{2024} \right)$$

Note that $$\frac{1}{2024} + \frac{1}{2024} = \frac{2}{2024} = \frac{1}{1012}$$.

So, $$\sum_{k=1}^{1012} \frac{1}{\alpha+k} = \frac{1}{1012} + \frac{1}{1013} + \dots + \frac{1}{2023}$$.

Comparing the terms:

• LHS starts at $$\frac{1}{\alpha+1}$$ and ends at $$\frac{1}{\alpha+1012}$$.

• RHS starts at $$\frac{1}{1012}$$ and ends at $$\frac{1}{2023}$$.

Matching the first terms: $$\alpha + 1 = 1012 \implies \mathbf{\alpha = 1011}$$