The problem involves evaluating the ratio test for the series:

$\sum_{n=1}^{\infty} \frac{n^2 + 3}{(n+1)!}$

Step-by-Step Solution:

1. Define the sequence $u_n$: The terms of the series are: $u_n = \frac{n^2 + 3}{(n+1)!}.$

2. Compute $\frac{u_{n+1}}{u_n}$: Substitute $u_{n+1} = \frac{(n+1)^2 + 3}{(n+2)!}$ and divide by $u_n$: $\frac{u_{n+1}}{u_n} = \frac{\frac{(n+1)^2 + 3}{(n+2)!}}{\frac{n^2 + 3}{(n+1)!}} = \frac{(n+1)^2 + 3}{n^2 + 3} \cdot \frac{(n+1)!}{(n+2)!}.$

3. Simplify the factorial term: Using the property $(n+2)! = (n+2)(n+1)!$, simplify: $\frac{u_{n+1}}{u_n} = \frac{(n+1)^2 + 3}{n^2 + 3} \cdot \frac{1}{n+2}.$

4. Take the limit as $n \to \infty$: For large $n$, the dominant term in the numerator and denominator is the squared term: $\frac{(n+1)^2 + 3}{n^2 + 3} \sim \frac{(n+1)^2}{n^2} = \frac{n^2 + 2n + 1}{n^2} = 1 + \frac{2}{n} + \frac{1}{n^2}.$ As $n \to \infty$, this approaches 1. The $\frac{1}{n+2}$ term vanishes in the limit: $\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = 0.$

5. Conclusion: Since the ratio limit is $0$, the series converges, and the ratio is finite.

Do you want further clarification on any step, or do you have questions?

Related Questions:

1. What is the ratio test and how is it used for series convergence?
2. How does factorial growth dominate polynomial growth in sequences?
3. Why does $\lim_{n \to \infty} \frac{(n+1)^2}{n^2} \to 1$?
4. What is the difference between absolute and conditional convergence of a series?
5. Can this series be evaluated using other convergence tests, such as the root test?

Tip: When dealing with factorials in series, simplify them using the property $(n+1)! = (n+1) \cdot n!$ to avoid unnecessary complexity.