1. **State the problem:** Evaluate the sum $$\frac{4}{2! + 3! + 4!} + \frac{5}{3! + 4! + 5!} + \frac{6}{4! + 5! + 6!} + \cdots + \frac{2025}{2023! + 2024! + 2025!}$$ 2. **Rewrite the denominator:** Given the hint, consider the general term for $n \geq 4$: $$\frac{n}{(n-2)! + (n-1)! + n!}$$ 3. **Factor the denominator:** Note that $$(n-1)! = (n-1)(n-2)!$$ $$n! = n (n-1)! = n (n-1)(n-2)!$$ So, $$(n-2)! + (n-1)! + n! = (n-2)! + (n-1)(n-2)! + n (n-1)(n-2)!$$ $$= (n-2)! \left[1 + (n-1) + n(n-1)\right]$$ 4. **Simplify the bracket:** Calculate $$1 + (n-1) + n(n-1) = 1 + n - 1 + n^2 - n = n^2$$ 5. **Rewrite denominator:** Therefore, $$(n-2)! + (n-1)! + n! = (n-2)! \cdot n^2$$ 6. **Rewrite the term:** The term becomes $$\frac{n}{(n-2)! \cdot n^2} = \frac{n}{n^2 (n-2)!} = \frac{1}{n (n-2)!}$$ 7. **Express in terms of factorials:** Recall that $$n (n-2)! = n! / (n-1)$$ because $$n! = n (n-1) (n-2)! \Rightarrow n (n-2)! = \frac{n!}{n-1}$$ So, $$\frac{1}{n (n-2)!} = \frac{1}{\frac{n!}{n-1}} = \frac{n-1}{n!}$$ 8. **Rewrite the sum:** The sum is $$\sum_{n=4}^{2025} \frac{n-1}{n!} = \sum_{n=4}^{2025} \frac{n-1}{n!}$$ 9. **Split the numerator:** $$\frac{n-1}{n!} = \frac{n}{n!} - \frac{1}{n!} = \frac{1}{(n-1)!} - \frac{1}{n!}$$ 10. **Rewrite the sum using this:** $$\sum_{n=4}^{2025} \left( \frac{1}{(n-1)!} - \frac{1}{n!} \right) = \sum_{n=4}^{2025} \frac{1}{(n-1)!} - \sum_{n=4}^{2025} \frac{1}{n!}$$ 11. **Change index in the first sum:** Let $k = n-1$, then when $n=4$, $k=3$, and when $n=2025$, $k=2024$: $$\sum_{k=3}^{2024} \frac{1}{k!} - \sum_{n=4}^{2025} \frac{1}{n!}$$ 12. **Combine sums:** $$\left( \frac{1}{3!} + \frac{1}{4!} + \cdots + \frac{1}{2024!} \right) - \left( \frac{1}{4!} + \frac{1}{5!} + \cdots + \frac{1}{2025!} \right)$$ 13. **Cancel common terms:** All terms from $\frac{1}{4!}$ to $\frac{1}{2024!}$ cancel out, leaving $$\frac{1}{3!} - \frac{1}{2025!}$$ 14. **Calculate the final value:** $$\frac{1}{3!} = \frac{1}{6}$$ Since $\frac{1}{2025!}$ is extremely small, the sum is approximately $$\frac{1}{6} - \frac{1}{2025!} \approx \frac{1}{6}$$ **Final answer:** $$\boxed{\frac{1}{6} - \frac{1}{2025!}}$$ This is the exact value of the sum.