Using the Cauchy-Hadamard Theorem to find a radius of convergence 
Cauchy-Hadamard Theorem 
Consider the formal power series $$f(z) = \sum_{n = 0}^{\infty}
c_{n}(z - a)^{n} $$
for $a, c_{n} \in \mathbb{C}$. Then the radius of convergence of $f$
  at the point $a$ is given by 
$$\frac{1}{R} = \limsup_{n\to\infty} (|c_{n}|^{1/n}).$$

Can I use this Theorem on this series:
$$\sum_{n \text{ even}}^{\infty} \frac{(-1)^{(n - 1)/2} }{\prod_{i = 1}^{(n - 1)/2} 2i + 1} \cdot x^{n}$$
I tried like this:
$$\frac{1}{R} = \lim_{n\to\infty} \left|\frac{(-1)^{(n - 1)/2}}{\prod_{i = 1}^{(n - 1)/2} 2i + 1} \right|^{1/n}$$
Let the limit equal $L$. Then
$$\text{log}(L) = \lim_{n\to\infty} \frac{1}{n} \log\left(\frac{(-1)^{(n - 1)/2}}{\prod_{i = 1}^{(n - 1)/2} 2i + 1} \right) = 0 $$
So $L = e^{0} = 1$ and $R = 1$. Is this right? I'm pretty sure the right answer is $R = \infty$, though. Does it not work because my sum is only going through even $n$?

\begin{align}|c_n| &= \begin{cases} \left|\frac{(-1)^{(n - 1)/2}}{\prod_{i = 1}^{(n - 1)/2} 2i + 1} \right|^{1/n} & \text{ if } n \text{ is even} \\0 &  \text{ if } n \text{ is odd}\end{cases} \\
&=\begin{cases} \left|\frac{1}{\prod_{i = 1}^{(n - 1)/2} 2i + 1} \right|^{1/n} & \text{ if } n \text{ is even} \\0 &  \text{ if } n \text{ is odd}\end{cases} \\\end{align}
and hence we have \begin{align}\lim\sup |c_n| &= \lim_{n \to \infty} \left|\frac{1}{\prod_{i = 1}^{(n - 1)/2} (2i + 1)} \right|^{1/n} \\
&=\lim_{n \to \infty} \left|\frac{n!2^n}{(2n)!} \right|^{1/n} \\
&=2 \lim_{n \to \infty} \frac{n^\frac1{2n}(n/e)}{(2n)^\frac1{2n}(2n/e)^2}\\
&=\frac{e}2 \lim_{n \to \infty}\frac{1}{n }\\&= 0\end{align}
where I have used Stirling approximation.
