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

1. Problem Description

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

2. Solution Steps

We use the Ratio Test. Let $a_n = \frac{n^3}{(2n)!}$. Then

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

We have $(2n+2)! = (2n+2)(2n+1)(2n)!$, so

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

Then

$$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left(\frac{n+1}{n}\right)^3 \cdot \frac{1}{(2n+2)(2n+1)} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^3 \cdot \frac{1}{4n^2+6n+2}$$

$$= (1)^3 \cdot \lim_{n\to\infty} \frac{1}{4n^2+6n+2} = 1 \cdot 0 = 0$$

Since the limit is 0, which is less than 1, the series converges by the Ratio Test.

3. Final Answer

The series converges.