The comparison test helps determine the convergence or divergence of a series by comparing it to another series that is known to converge or diverge. To use the comparison test on

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

we need to compare this series to a simpler one whose convergence properties are well-known. Here’s a step-by-step analysis:

1. Simplify the Terms: Start by simplifying the expression $\frac{n!}{(n+2)!}$.

   • Note that $(n+2)! = (n+2)(n+1)n!$, so we can rewrite $\frac{n!}{(n+2)!}$ as: $\frac{n!}{(n+2)(n+1)n!} = \frac{1}{(n+2)(n+1)}.$

2. Form of the Simplified Series: This means the series becomes: $\sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)}.$

3. Comparison with a Known Series: Now, let’s compare $\frac{1}{(n+2)(n+1)}$ to a simpler form that we can analyze. Notice that: $\frac{1}{(n+2)(n+1)} < \frac{1}{n^2}.$

   Since the series $\sum \frac{1}{n^2}$ is a convergent $p$-series (with $p = 2 > 1$), and $\frac{1}{(n+2)(n+1)}$ is smaller than $\frac{1}{n^2}$ for large $n$, we can conclude that our original series $\sum \frac{n!}{(n+2)!}$ converges by the comparison test.

Conclusion:

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

Would you like a more detailed explanation or have any other questions?

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Here are some related questions to deepen understanding:

1. What are some other convergence tests that can be used on factorial-based series?
2. How does the limit comparison test differ from the comparison test, and can it be applied here?
3. Could the ratio test also determine convergence for this series?
4. What are $p$-series, and why are they useful in convergence analysis?
5. How can factorial simplification techniques assist in evaluating series convergence?

Tip: For factorial expressions in series, try simplifying the ratio of consecutive terms to make convergence easier to analyze.