Step 1: Understanding the Topic
This question asks for the radius of convergence of a power series. The presence of factorial terms and powers of $n$ suggests that the Ratio Test is the most effective method for determining the convergence properties of the series.
Step 2: Key Formula or Approach
We use the Ratio Test. For the series $\sum a_n x^n$, where $a_n = \frac{\binom{n}{6}^2}{(2n)!}$, we need to compute the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. The radius of convergence is then $R = 1/L$. (If $L=0$, $R=\infty$; if $L=\infty$, $R=0$).
Step 3: Detailed Calculation
A. Set up the ratio:
\[
\frac{a_{n+1}}{a_n} = \frac{\binom{n+1}{6}^2}{(2n+2)!} \cdot \frac{(2n)!}{\binom{n}{6}^2} = \left(\frac{\binom{n+1}{6}}{\binom{n}{6}}\right)^2 \cdot \frac{(2n)!}{(2n+2)!}
\]
B. Simplify each part of the ratio:
First, the ratio of binomial coefficients:
\[
\frac{\binom{n+1}{6}}{\binom{n}{6}} = \frac{\frac{(n+1)!}{6!(n-5)!}}{\frac{n!}{6!(n-6)!}} = \frac{(n+1)!}{n!} \cdot \frac{(n-6)!}{(n-5)!} = (n+1) \cdot \frac{1}{n-5} = \frac{n+1}{n-5}
\]
Second, the ratio of factorials:
\[
\frac{(2n)!}{(2n+2)!} = \frac{(2n)!}{(2n+2)(2n+1)(2n)!} = \frac{1}{(2n+2)(2n+1)}
\]
C. Combine and take the limit:
The full ratio is:
\[
\left|\frac{a_{n+1}}{a_n}\right| = \left(\frac{n+1}{n-5}\right)^2 \cdot \frac{1}{(2n+2)(2n+1)}
\]
Now, we find the limit as $n \to \infty$:
\[
L = \lim_{n \to \infty} \left[ \left(\frac{n+1}{n-5}\right)^2 \cdot \frac{1}{(2n+2)(2n+1)} \right]
\]
The first part of the limit is:
\[
\lim_{n \to \infty} \left(\frac{n+1}{n-5}\right)^2 = \left(\lim_{n \to \infty} \frac{1+1/n}{1-5/n}\right)^2 = (1)^2 = 1
\]
The second part of the limit has a denominator that grows like $4n^2$, so it goes to zero.
\[
\lim_{n \to \infty} \frac{1}{(2n+2)(2n+1)} = 0
\]
Therefore, the overall limit is:
\[
L = 1 \times 0 = 0
\]
D. Find the Radius of Convergence:
Since $L=0$, the radius of convergence $R = 1/L$ is infinite.
Step 4: Final Answer
The series converges for all real numbers $x$.
\[
\boxed{R = \infty}
\]