How do I find the radius of convergence of this powerseries? How do I find the radius of convergence, $R$, for this series?
$$\sum_{n=2}^{\infty} \frac{(-1)^n z^{2n}}{(2n-1)!} $$
I know it is a power series and I would think I am supposed to find the radius of convergence from the formula: 
$$R = \lim_{n \rightarrow \infty}\frac{|a_n|}{|a_{n+1}|}$$
The problem I see here is that the sum does not start from $0$ and that the exponent of $z$ is not $n$. How does these things affect the radius of convergence and is it still possible to use this approach if I rewrite the sum in some way. 
 A: For your first question, 
If a general power series is of the form $\sum_{n=0}^{\infty}a_nz^n$ then for your power series:$$a_0=a_{2n+1}=0\text{ for all }n\geq0\\a_{2n}=\frac{(-1)^n}{(2n-1)!}\text{ for all } n\geq1$$
For your second question, 
$$\lim \sup |a_n|^{\frac{1}{n}}=\lim \sup |a_{2n}|^{\frac1{2n}}=0$$
Thus, $R =\infty.$
A: HINT:
$$S=\sum_{n=2}^{\infty} \frac{(-1)^n z^{2n}}{(2n-1)!}=\sum_{n=2}^{\infty}\dfrac{(-z^2)^n}{(2n-1)!}$$
Now if $\dfrac{(-z^2)^n}{(2n-1)!}=f(n), f(0)=0, f(1)=?$
$$\implies S=-\dfrac{(-z^2)}{1!}+\sum_{n=0}^{\infty}\dfrac{(-z^2)^n}{(2n-1)!}$$
A: First, since you are taking a limit as $n\to \infty$, you can safely ignore first finitely many terms.  In particular, the "missing" first two terms won't change the RoC.
Second, you can rewrite the series so that it is a power series in terms of some complex variable raised to the $n$.  You could write
$$ \sum_{n=2}^{\infty} \frac{(-1)^n z^{2n}}{(2n-1)!} = \sum_{n=2}^{\infty} \frac{(-1)^n w^n}{(2n-1)!} $$
where $w = z^2$.  Compute the RoC for the new series (in $w$), then use that to determine conditions on $z$.
Can you finish it?
A: The coefficient $a_n$ of $z^n$ is $a_n=\frac{(-1)^{n/2}}{(n-1)!}$ if $n$ is even and $\ge 4$, and is $a_n=0$ otherwise.
We have 
$$\limsup_{n\to \infty}\sqrt[n]{|a_n|}=0 $$
and hence convergence radius $R=\infty$. Note that neither the finitely many exceptions for small $n$ nor the infinitely many zero coefficients for odd $n$ are relevant in computing the limsup.
