Question 3 Find the radius $R$ and the interval of convergence $I O C$ of the series. \[ \sum_{n=1}^{\infty} \frac{n^{2} x^{n}}{2 \cdot 4 \cdot 6 \cdots(2 n)} \] $R=5, \quad I O C=(-4,6)$ $R=5, \quad I O C=[-5,5]$ $R=\infty, \quad I O C=(-\infty, \infty)$

Step 1 ：First, we need to find the general term of the series. The general term is given by \(a_n = \frac{n^{2} x^{n}}{2 \cdot 4 \cdot 6 \cdots(2 n)}\).

Step 2 ：We can simplify this to \(a_n = \frac{n^{2} x^{n}}{2^n n!}\).

Step 3 ：Next, we need to find the ratio of successive terms, \(\frac{a_{n+1}}{a_n}\).

Step 4 ：Calculating this gives \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^2 x^{n+1}}{2^{n+1} (n+1)!} \cdot \frac{2^n n!}{n^2 x^n} = \frac{(n+1)x}{2n}\).

Step 5 ：As n approaches infinity, the ratio \(\frac{(n+1)x}{2n}\) approaches \(\frac{x}{2}\).

Step 6 ：The series converges when the absolute value of the ratio is less than 1, i.e., \(|\frac{x}{2}| < 1\).

Step 7 ：Solving this inequality gives the interval of convergence as \(-2 < x < 2\).

Step 8 ：The radius of convergence R is the half-length of the interval of convergence, so \(R = 2\).

Step 9 ：Finally, we need to check the endpoints of the interval of convergence. Substituting \(x = -2\) and \(x = 2\) into the original series, we find that the series diverges at both endpoints.

Step 10 ：So, the interval of convergence is \((-2, 2)\) and the radius of convergence is \(\boxed{2}\).