To solve this problem, we will apply the ratio test to show that:

$\lim_{n \to \infty} \frac{(2n)!}{3^{2n} (n!)^2} = 0$

Step 1: Apply the Ratio Test

Let $a_n = \frac{(2n)!}{3^{2n} (n!)^2}$. According to the ratio test, we need to compute the limit:

$L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$

We now compute $\frac{a_{n+1}}{a_n}$:

$\frac{a_{n+1}}{a_n} = \frac{\frac{(2(n+1))!}{3^{2(n+1)} ((n+1)!)^2}}{\frac{(2n)!}{3^{2n} (n!)^2}}$

Simplifying this:

$\frac{a_{n+1}}{a_n} = \frac{(2(n+1))! \cdot 3^{2n} \cdot (n!)^2}{(2n)! \cdot 3^{2(n+1)} \cdot ((n+1)!)^2}$

This simplifies further:

$\frac{a_{n+1}}{a_n} = \frac{(2n+2)(2n+1)(2n)! \cdot (n!)^2}{(2n)! \cdot 9 \cdot 3^{2n} \cdot (n+1)^2 \cdot (n!)^2}$

$= \frac{(2n+2)(2n+1)}{9(n+1)^2}$

Step 2: Take the Limit

Now, compute the limit as $n \to \infty$:

$L = \lim_{n \to \infty} \frac{(2n+2)(2n+1)}{9(n+1)^2}$

Dividing both numerator and denominator by $n^2$:

$L = \lim_{n \to \infty} \frac{4n^2 + 6n + 2}{9(n^2 + 2n + 1)} = \frac{4}{9}$

Since $\frac{4}{9} < 1$, the ratio test confirms that the series converges.

Step 3: Conclusion

By the ratio test, the given expression converges to 0 as $n \to \infty$.

Do you need more details, or would you like to explore another approach?

Here are five related questions you might find helpful:

1. What other convergence tests could be used to solve similar problems?
2. How does the ratio test apply to series with factorial terms in general?
3. What is the significance of the ratio test result $L < 1$?
4. How can Stirling's approximation be used to approximate factorials in such limits?
5. What other limits can be evaluated using the ratio test?

Tip: When dealing with factorial expressions, consider applying Stirling's approximation for large $n$ to simplify the calculations.