Step 1: Understanding the Topic
This question asks for the radius of convergence of a power series. A power series converges within a certain interval, and the radius of convergence defines the size of this interval. The most common method to find this radius is the Ratio Test.
Step 2: Key Formula or Approach
We will use the Ratio Test for power series. For a series $\sum a_n x^n$, we compute the limit:
\[
L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|
\]
The series converges if $L|x|<1$, which means $|x|<1/L$. The radius of convergence is $R = 1/L$.
Here, the coefficient is $a_n = \frac{(n!)^2}{(2n)!}$.
Step 3: Detailed Calculation
A. Set up the ratio $\left|\frac{a_{n+1}{a_n}\right|$:}
\[
\left|\frac{a_{n+1}}{a_n}\right| = \frac{((n+1)!)^2}{(2(n+1))!} \div \frac{(n!)^2}{(2n)!} = \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2}
\]
B. Simplify the factorial expressions:
We use the properties $(n+1)! = (n+1)n!$ and $(2n+2)! = (2n+2)(2n+1)(2n)!$.
\[
\frac{((n+1)n!)^2}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{(n!)^2}
\]
Cancel the $(n!)^2$ and $(2n)!$ terms:
\[
= \frac{(n+1)^2}{(2n+2)(2n+1)} = \frac{(n+1)^2}{2(n+1)(2n+1)} = \frac{n+1}{2(2n+1)} = \frac{n+1}{4n+2}
\]
C. Compute the limit L:
\[
L = \lim_{n\to\infty} \frac{n+1}{4n+2}
\]
To evaluate this limit, we can divide the numerator and denominator by the highest power of $n$:
\[
L = \lim_{n\to\infty} \frac{1 + 1/n}{4 + 2/n} = \frac{1+0}{4+0} = \frac{1}{4}
\]
D. Find the Radius of Convergence R:
The radius of convergence is $R = \frac{1}{L}$.
\[
R = \frac{1}{1/4} = 4
\]
The series converges for $|x|<4$.
Step 4: Final Answer
The radius of convergence is 4.
\[
\boxed{R = 4}
\]