We are given the sum:

$S = \frac{1}{2 \left(1 + \frac{1}{2}\right)} + \frac{1}{3 \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right)} + \cdots + \frac{1}{2025 \left(1 + \frac{1}{2}\right) \left(1 + \frac{1}{3}\right) \cdots \left(1 + \frac{1}{2025}\right)}$

Step 1: Simplify the general term

Each term in the sum can be written in the form:

$T_n = \frac{1}{n \prod_{k=2}^{n} \left(1 + \frac{1}{k}\right)}$

The term $1 + \frac{1}{k}$ simplifies to $\frac{k+1}{k}$, so:

$T_n = \frac{1}{n \prod_{k=2}^{n} \frac{k+1}{k}} = \frac{1}{n \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdots \frac{n+1}{n}}$

Notice that most terms in the product cancel out, leaving:

$T_n = \frac{1}{n \cdot \frac{n+1}{2}} = \frac{2}{n(n+1)}$

Step 2: Rewrite the sum

Thus, the sum can be rewritten as:

$S = \sum_{n=2}^{2025} \frac{2}{n(n+1)}$

Step 3: Simplify the sum using partial fractions

We can break $\frac{2}{n(n+1)}$ into partial fractions:

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

Step 4: Use telescoping series

Now, the sum becomes:

$S = \sum_{n=2}^{2025} \left( \frac{2}{n} - \frac{2}{n+1} \right)$

This is a telescoping series, where most of the terms cancel out, leaving:

$S = \left( \frac{2}{2} - \frac{2}{2026} \right) = 1 - \frac{2}{2026}$

Final Answer:

$S = 1 - \frac{1}{1013}$

Thus, the value of the sum is:

$S = \frac{1012}{1013}$

Would you like further clarification or a detailed breakdown of any part?

Here are some related questions:

1. What is a telescoping series and how does it work?
2. How can you find partial fraction decompositions for more complicated expressions?
3. How do you recognize patterns in sums or series to simplify them?
4. Can this approach be applied to other types of series, like geometric or harmonic series?
5. How can we apply these techniques to approximate sums or integrals?

Tip: In telescoping series, try to recognize when terms cancel each other out, which dramatically simplifies the computation!