Question

Find the radius of convergence of the series
\[ \sum_{n=0}^{\infty} \frac{\binom{n}{6}^2}{(2n)!} x^n \]

Hint:

Factorials grow much faster than polynomial expressions. If factorial growth dominates in the denominator, the radius of convergence is often infinite.

Answer 1

We use the Ratio Test to find the radius of convergence.
Let \[ a_n = \frac{\binom{n}{6}^2}{(2n)!} \] Then the given series is \[ \sum a_n x^n \] Step 1: Apply the Ratio Test.
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x| \] Now, \[ \frac{a_{n+1}}{a_n} = \frac{\binom{n+1}{6}^2}{(2n+2)!} \cdot \frac{(2n)!}{\binom{n}{6}^2} \] Step 2: Simplify the binomial ratio.
For large \( n \), \[ \binom{n}{6} = \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{6!} \] So asymptotically, \[ \binom{n}{6} \sim \frac{n^6}{6!} \] Thus, \[ \binom{n}{6}^2 \sim C n^{12} \] for some constant \( C \).
Step 3: Use factorial growth comparison.
We know \[ (2n+2)! = (2n+2)(2n+1)(2n)! \] So, \[ \frac{(2n)!}{(2n+2)!} = \frac{1}{(2n+2)(2n+1)} \] As \( n \to \infty \), \[ (2n+2)(2n+1) \sim 4n^2 \] Thus, \[ \frac{a_{n+1}}{a_n} \sim \frac{(n+1)^{12}}{n^{12}} \cdot \frac{1}{4n^2} \] As \( n \to \infty \), \[ \frac{(n+1)^{12}}{n^{12}} \to 1 \] Hence, \[ \frac{a_{n+1}}{a_n} \sim \frac{1}{4n^2} \] Step 4: Take the limit.
\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \] Thus, \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| |x| = 0 \] for every finite \( x \).
Conclusion:
Since the limit is zero for all \( x \), the series converges for every real number.
Final Answer: \[ \boxed{R = \infty} \]