Question

If $\left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \dots + \frac{1}{\alpha + 1012} \right) - \left( \frac{1}{2 \cdot 1} + \frac{1}{4 \cdot 3} + \frac{1}{6 \cdot 5} + \dots + \frac{1}{2024 \cdot 2023} \right)$

Answer 1

Step 1: First Sum Expression

We are given:

$S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha + k}$

This is the sum of the reciprocals of consecutive integers starting from $\alpha + 1$ to $\alpha + 1012$.

Step 2: Second Sum Expression

The second sum is:

$S_2 = \sum_{k=1}^{1012} \frac{1}{2k \cdot (2k - 1)}$

We can simplify each term of the second sum:

$\frac{1}{2k \cdot (2k - 1)} = \frac{1}{2k - 1} - \frac{1}{2k}$

Thus, the second sum $S_2$ becomes:

$S_2 = \sum_{k=1}^{1012} \left( \frac{1}{2k - 1} - \frac{1}{2k} \right)$

This is a telescoping series, and many terms cancel out. After cancellation, we are left with:

$S_2 = 1 - \frac{1}{2024}$

Step 3: Combine the Two Sums

Now, we substitute both sums $S_1$ and $S_2$ into the given equation:

$S_1 - S_2 = \frac{1}{2024}$

Substitute $S_2$:

$S_1 - \left( 1 - \frac{1}{2024} \right) = \frac{1}{2024}$

Simplifying the equation:

$S_1 - 1 + \frac{1}{2024} = \frac{1}{2024}$

$S_1 = 1$

Step 4: Solving for $\alpha$

We know that:

$S_1 = \sum_{k=1}^{1012} \frac{1}{\alpha + k} = 1$

We can now express the equation as:

$\left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} \right) = \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2023} \right) + \frac{1}{2024}$

Rewriting this as:

$\left( \frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} \right) = \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2023} \right) + \frac{1}{2024}$

This simplifies to:

$\frac{1}{\alpha + 1} + \frac{1}{\alpha + 2} + \cdots + \frac{1}{\alpha + 1012} = 1 + 1 + \frac{1}{2024}$

The equation simplifies to:

$\alpha = 1011$

Thus, the value of $\alpha$ is: 1011