We need to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{n^3}{(2n)!}$ using the Ratio Test.

1. Problem Description

We need to determine the convergence or divergence of the series $\sum_{n=1}^{\infty} \frac{n^3}{(2n)!}$ using the Ratio Test.

2. Solution Steps

The Ratio Test states that for a series $\sum_{n=1}^{\infty} a_n$, we compute the limit $L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$.

If $L < 1$, the series converges absolutely.

If $L > 1$, the series diverges.

If $L = 1$, the test is inconclusive.

In this case, $a_n = \frac{n^3}{(2n)!}$. Thus, $a_{n+1} = \frac{(n+1)^3}{(2(n+1))!} = \frac{(n+1)^3}{(2n+2)!}$.

We need to find the limit $L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n\to\infty} \left| \frac{\frac{(n+1)^3}{(2n+2)!}}{\frac{n^3}{(2n)!}} \right|$.

$L = \lim_{n\to\infty} \left| \frac{(n+1)^3}{(2n+2)!} \cdot \frac{(2n)!}{n^3} \right| = \lim_{n\to\infty} \frac{(n+1)^3}{n^3} \cdot \frac{(2n)!}{(2n+2)!}$.

We know that $\frac{(n+1)^3}{n^3} = \left( \frac{n+1}{n} \right)^3 = \left( 1 + \frac{1}{n} \right)^3$.

As $n \to \infty$, $\left( 1 + \frac{1}{n} \right)^3 \to (1+0)^3 = 1$.

Also, $\frac{(2n)!}{(2n+2)!} = \frac{(2n)!}{(2n+2)(2n+1)(2n)!} = \frac{1}{(2n+2)(2n+1)}$.

So, $L = \lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^3 \cdot \frac{1}{(2n+2)(2n+1)} = \lim_{n\to\infty} 1 \cdot \frac{1}{4n^2 + 6n + 2} = 0$.

Since $L = 0 < 1$, the series converges absolutely by the Ratio Test.

3. Final Answer

The series converges.