Find radius of convergence of power series: $\sum_{n=0}^\infty (-1)^n z^{2n} / (2n)!$ We have the power series $ \sum\nolimits_{n=1}^\infty \frac{(-1)^n}{(2n)!} z^{2n}$. How do I get the $c_n$ to use the formula: $C= \lim\limits_{n \rightarrow \infty}\frac{|c_n|}{|c_{n+1}|}$ ?
I don't know if I can use $\frac{(-1)^n}{(2n)!}$ as $c_n$ because $z$ has $2n$ in the exponent.
Can I transform the series to this form: $ \sum\nolimits_{n=1}^\infty c_n(z-a)^n$ ?
 A: Use d'Alembert classical ratio  test.
Let $u_n(z)=\frac {(-1)^nz^{2n}}{(2n)!} $.
$$\lim_{+\infty}\frac {|u_{n+1}(z)|}{|u_n(z)|}=$$
$$=\lim_{+\infty}\frac {|z|^2}{(2n+1)(2n+2)}=0$$
the series is absolutely convergent for all complex numbers $z $.
the radius is infinite.
$$\sum_{n=0}^{+\infty}u_n(z)=\cos (z) $$
A: Let $z^2=y .$ Then by the ratio test the series converges for all $y,$ hence for all $z.$
A: It is also straightforward by "definition" of the radius of convergence, since $\limsup\sqrt[n]{|c_n|}=\lim \frac{1}{\sqrt[2n]{(2n)!}}$, and $\sqrt[2n]{(2n)!} \to + \infty$, which can be seen by the inequality
\begin{align*}
\sqrt[2n]{(2n)!} &\geq \sqrt[2n]{\left(\frac{2n}{2}\right)^n} \\
&=n^{1/2}.
\end{align*}
A: You are correct that the "shortcut" method isn't safe to use.
Simple example of it failing:
$\displaystyle \sum_{n=0}^{+\infty} \left(\frac x2\right)^{2n} = \sum_{n=0}^{+\infty} \left(\frac{x^2}4\right)^n$ is a geometric series which converges for $x^2/4 < 1$, i.e., for $|x| < 2$.
If we rewrite the series as $\displaystyle \sum_{n=0}^{+\infty} \frac1{4^n} x^{2n}$ then we can try using the "shortcut" method with $c_n = \dfrac 1{4^n}$ and $c_{n+1} = \dfrac1{4^{n+1}}$.  This will give us $$C = \lim_{n\to+\infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n\to+\infty} \frac{4^{n+1}}{4^n} = 4.$$
This means the radius of convergence is $4$, but this is most certainly not the case since $|x| < 2$ means the radius of convergence is $2$.
Now, this doesn't mean that the "shortcut" method will always fail when the exponent on the variable is $an$ for some $a \ne 1$.  It just means that it doesn't always work, and if it doesn't always work for a certain case, then don't use it in those cases.  It's better to use something that we know works 100% of the time in those cases.  And in this case, the thing to use is the standard ratio test.
For your specific example, let $a_n = \dfrac{(-1)^n}{(2n)!} z^{2n}$.  Then find $a_{n+1}$ and then find the values of $z$ that satisfy $\displaystyle\lim_{n\to+\infty} \left|\frac{a_{n+1}}{a_n}\right| < 1.$  (This was worked out in the other answer.)  I also encourage you to try the "shortcut" method on this to see what happens.

Can I transform the series to this form: $ \sum\nolimits_{n=1}^\infty c_n(z-a)^n$ ?

Not in a useful way.  If you do this then you'll have $a = 0$ (which is fine), and $$c_n = \begin{cases} 1/n!, & \text{if } n \bmod 4 = 0, \\ -1/n!, & \text{if } n \bmod 4 = 2, \\ 0 & \text{if $n$ is odd}. \end{cases}$$
This form of $c_n$ is not helpful.
A: The ratio test handles this:
\begin{align}
& \lim_{n\to\infty}\frac{\left| \dfrac{(-1)^{n+1} z^{2(n+1)}}{(2(n+1))!} \right|}{\left| \dfrac{(-1)^n z^{2n}}{(2n)!} \right|} = \lim_{n\to\infty} \frac{(2n)!|z^2|}{(2(n+1))!} \\[12pt]
={} & |z^2| \lim_{n\to\infty} \frac{(2n)!}{(2n+2)!} \quad \text{This step is posssible because $|z^2|$ does not change as $n$  changes.} \\[12pt]
= {} & |z^2| \lim_{n\to\infty} \frac{(2n)!}{(2n)!(2n+1)(2n+2)} = |z^2| \lim_{n\to\infty} \frac 1 {(2n+1)(2n+2)} \\[12pt]
= {} & 0 \text{ regardless of what number $|z^2|$ is.}
\end{align}
Therefore the series converges regardless of how big $|z|$ is. The radius of convergence is infinite.
