Radius of Convergence and its application to a Power Series including $x^{2n}$ rather than $x^n$ 
(Radius of Convergence) Consider the Power Series $f(x)=\sum_{n=0}^{+ \infty}a_n x^n$, the radius of convergence $\rho$ can be found using $$\rho = \displaystyle \lim_{n \to + \infty} \left| \frac{a_n}{a_{n+1}} \right|$$

I did quite a few exercises already using the above definition and successfully computed the radius of convergence with it. But I stumbled across this example which I find interesting:
$$\sum_{n=0}^{+ \infty} \frac{(2n)!!}{n^n}x^{2n}\tag{taken from Michaels Analysis I}$$
For the sake of the argument let $0^0:=1$, otherwise just start from $n=1$.
First I note that this, from a narrow point of view, is a power series to me. However, I can see that it  is not quite in the form of $\sum a_n x^n$ but in the form $\sum a_n x^{2n}$. My leading impulse was to ignore that fact and just compute it anyway using the above definition. Here is what came out:
$$\displaystyle \lim_{n \to + \infty} \frac{(2n)!!}{n^n}\cdot \frac{(n+1)^{n+1}}{(2n+2)!!}= \lim_{n \to + \infty} \frac{(2n)!!(n+1)}{(2n+2)(2n)!!} \cdot \left(\frac{n+1}{n}\right)^n= \frac{e}{2}=\rho$$
so I would have concluded that $|x|< \frac{e}{2}$ is the Radius of convergence (without having checked the behavior on the endpoints). This is not the right answer, the correct answer would be $\rho = \sqrt{ e / 2}$.
Now as I looked for similar exercises on this website I often saw that a lot of people don't use the above definition and just use the Quotient Criteria (d'Alembert Criteria) to find the Radius of convergence. Indeed, using this criteria here things work out much nicer: $$ \lim_{n \to + \infty} \left|\frac{a_{n+1}}{a_n} \right|=\lim_{n \to + \infty} \frac{(2n+2)!!x^{2n+2}}{(n+1)^{n+1}}\cdot \frac{n^n}{(2n)!!x^{2n}}=\lim_{n \to + \infty} x^2\frac{(2n+2)(2n)!!}{(n+1)(2n)!!}\left( \frac{n}{n+1}\right)^n \\ = \frac{2x^2}{e}\overset{!}<1 \implies x^2 < \frac{e}{2}\implies \rho = \sqrt{\frac{e}{2}}$$
Question: Is there a specific way of route I can go into to decide how to find the radius of convergence? It might be a bit far fetched to say such a thing, but what I have seen on this site, most people use the Quotient Criteria rather then the definition above.
Also, is there a way I could alter the above example to use the Radius of Convergence definition I have given above in the header? Or would this be too much trouble in the first place to even consider doing such?
Additional: In the same book, C.T. Michaels Analysis 1, he shows that the Radius of Convergence for the $J_0$ Besselfunction is $\infty$ by using the exact definition as in the header above: $$J_0(x)= \sum_{n=0}^{+ \infty} \frac{(-1)^nx^{2n}}{2^{2n}(n!)^2}$$
which might help you to understand or rather highlight my confusion. 
 A: Consider a power series of the form $\sum_{n}a_n x^{2n}$, and note that this equals $\sum_{n}b_n x^{n}$, where $b_n=a_{n/2}$ for even $n$ and $b_n=0$ for odd $n$.  The ratio-based equality you give,
$$
\rho = \lim_{n\rightarrow\infty}\left|\frac{b_n}{b_{n+1}}\right|,
$$
only applies when the limit exists.  In this case (since every other coefficient is zero), the limit does not exist, so this equality cannot be used.  Instead you must use the more general result that
$$
\rho=\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|b_n|}}
$$
(this is the Cauchy-Hadamard theorem).  The zeroes do not contribute to the $\limsup$ (since there are infinitely many nonzero entries), and so this just gives
$$
\rho=\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[2n]{|b_{2n}|}}=\sqrt{\frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}}}=\sqrt{\tilde\rho},
$$
where $\tilde\rho$ is the radius of convergence of $\sum_n a_n x^n$.  In your case, you calculated $\tilde\rho=e/2$ (using the ratio equality, since the limit of the ratios of the successive $a_n$ does exist), and can conclude that $\rho=\sqrt{e/2}$.
A: Use mjqxxxx already gave you the best and most general way to do this. However, for power series like this one, there is a small trick. You can substitute $y=x^2$, use the result you quote to find the radius of convergence as a power series in $y$, then take the square root of this to find the radius of convergence as a power series in $x$. 
Note that this is essentially what you did, and this explains why your answer is missing a square root.
