Radius of Convergence for $\sum \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n$ I'm trying to find the radius of convergence for this series:
$$\sum \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n$$
so I have,
$$R=\lim_{n\to\infty} \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}\frac{2^{2n+2}(2n+2)!}{[1\cdot 3 \cdots (2n+1)]^2}=\lim_{n\to\infty}\frac{4(2n+1)(2n+2)}{(2n)^2(2n+1)^2}=0$$
But the answer says $4$ and I'm not sure how to get that. Any help?

$$\left(1 \cdot 3 \cdots (2n-1)\right)^2 = \left(\dfrac{(2n)!}{2^n \cdot (n!)} \right)^2$$
Hence, we have
$$\sum_{n} \frac{[1\cdot 3 \cdots (2n-1)]^2}{2^{2n}(2n)!}x^n = \sum_{n} \frac{(2n)!^2}{2^{2n}2^{2n}(2n)! n! n!}x^n = \sum_n \dbinom{2n}n \left(\dfrac{x}{16} \right)^n$$
Can you now work it out?
