Question: Find the radius of convergence of the given power series.
$$\sum^\infty _{k=0} \frac{(k!)^2}{(2k)!}(z-2)^k$$
Answer: By ratio test, we have that:
$$\frac{1}{R} = lim_{k\rightarrow \infty} |\frac{\frac{((k+1)!)^2}{(2(k+1))!}}{\frac{(k!)^2}{(2k)!}}|$$
$$= lim_{k\rightarrow \infty} |\frac{\frac{(k+1)^2(k)^2(k-1)^2...}{(2k+2)(2k+1)(2k)...}}{\frac{(k)^2(k-1)^2...}{(2k)(2k-1)...}}|$$
$$= lim_{k\rightarrow \infty} |\frac{(k+1)^2}{(2k+2)(2k+1)}|$$
$$= lim_{k\rightarrow \infty} \frac{k+1}{4k+2}$$
By L’Hospital’s Rule, we know that
$$\frac{1}{R} = lim_{k\rightarrow \infty} \frac{k+1}{4k+2} = lim_{k\rightarrow \infty} \frac{1}{4} = \frac{1}{4}$$
Therefore, we conclude that $R=4$.
