Question

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

Hint:

For power series, always apply ratio test and include $|x|$ before taking limit.

Answer 1

Correct Answer: 4

Step 1: Use Ratio Test.
Let \[ a_n = \frac{(n!)^2}{(2n)!}. \]
Consider: \[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{((n+1)!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2}. \]
Step 2: Simplify factorial terms.
\[ (n+1)! = (n+1)n!, \] \[ (2n+2)! = (2n+2)(2n+1)(2n)!. \]
Substitute: \[ \frac{((n+1)n!)^2}{(2n+2)(2n+1)(2n)!} \cdot \frac{(2n)!}{(n!)^2}. \]
Cancel $(n!)^2$ and $(2n)!$:
\[ = \frac{(n+1)^2}{(2n+2)(2n+1)}. \]
Step 3: Take limit as $n \to \infty$.
For large $n$:
\[ \frac{(n+1)^2}{(2n+2)(2n+1)} \sim \frac{n^2}{4n^2} = \frac{1}{4}. \]
Thus, \[ \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{4}. \]
Step 4: Apply Ratio Test with $x^n$.
The ratio becomes: \[ \left|\frac{a_{n+1}x^{n+1}}{a_n x^n}\right| = \frac{1}{4}|x|. \]
For convergence: \[ \frac{1}{4}|x|<1. \]
\[ |x|<4. \]
Radius of Convergence:
\[ \boxed{R = 4}. \]