Difficulties evaluating the endpoints of the radius of convergence for a particular power series. I am having difficulties evaluating the endpoints of the radius of convergence for the following power series.
$$\sum_{k=0}^{\infty}\frac{(k!)^2 x^k}{(2k)!}$$
Using the ration test we get |x|<4. However, evaluating $x=4$ and $x=-4$ is proving immensely difficult. The solutions neglect this part, and my teacher is at loss. Any hints? 
In other words, I am having difficulty evaluating
$$\sum_{k=0}^{\infty}\frac{(k!)^2 4^k}{(2k)!}.$$
 A: Using Stirling's approximation
$$n! \sim \sqrt{2\pi n}\cdot \left(\frac{n}{e}\right)^n,$$
we obtain
$$\frac{(k!)^24^k}{(2k)!} \sim \frac{2\pi k\cdot k^{2k}e^{-2k}4^k}{\sqrt{2\pi(2k)}\cdot(2k)^{2k}e^{-2k}} = \sqrt{\pi k},$$
so the term of the series doesn't converge to $0$ for $\lvert x\rvert = 4$.
A: $\begin{align*}
a_k = \frac{(k!)^2 4^k}{(2k)!} &= \frac{(k!)^2 4^k}{(2k)(2k-1)(2k-2) \cdots 3\cdot 2 \cdot 1} \\
&=\frac{(k!)^2 4^k}{((2k)(2k-2) \cdots 4 \cdot 2)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{(k!)^2 4^k}{(2^k k!)((2k-1)(2k-3) \cdots 3 \cdot 1)} \\
&=\frac{k! \cdot 2^k}{(2k-1)(2k-3) \cdots 3 \cdot 1} \\
&=\frac{k! \cdot 2^k}{2\cdot(k-\frac{1}{2})\cdot 2\cdot(k-\frac{3}{2}) \cdots 2\cdot(\frac{3}{2}) \cdot 2\cdot \frac{1}{2}} \\
&=\frac{k! \cdot 2^k}{2^k(k-\frac{1}{2})(k-\frac{3}{2}) \cdots \cdot(\frac{3}{2}) (\frac{1}{2})} \\
&= \left( \frac{k}{k-\frac{1}{2}}\right)\left( \frac{k-1}{k-\frac{3}{2}}\right) \cdots \left(\frac{2}{\frac{3}{2}}\right)\left(\frac{1}{\frac{1}{2}}\right)\\
&\geq 1 \cdot 1 \cdots 1 \cdot 1 \\
&=1
\end{align*}$
When $x = 4$, $a_k\geq 1$, so $\displaystyle \lim_{k \rightarrow \infty} a_k \neq 0$, and the series diverges. Similarly true when $x=-4$.
