$J(z) = \sum^{\infty}_{k=0}\frac{(-1)^kz^{2k}}{(k!)^22^{2k}}$ should converge for all $z$ but doesn't. Why? 
Show that$$J(z) =
 \sum^{\infty}_{k=0}\frac{(-1)^kz^{2k}}{(k!)^22^{2k}}$$ converges for
  all $z \in C$ (can't figure out how to do a complex C).

I've managed to "prove" that $J(z)$ does not converge for all $z$ and I'm wondering where my "proof" is wrong.
Let $a_k = \frac{(-1)^kz^{2k}}{(k!)^22^{2k}}.$
A theorem states that if $p = \lim_{k\to\infty} |a_k|^{\frac{1}{k}} > 1$, then $J(z)$ diverges.
$$p = \lim_{k\to\infty} |a_k|^{\frac{1}{k}} = \left|\frac{(-1)^kz^{2k}}{(k!)^22^{2k}}\right|^{\frac{1}{k}} = \lim_{k\to\infty} \left|\frac{-z^2}{((k!)^\frac{1}{k})^2\cdot4}\right| = \lim_{k\to\infty}\frac{|z|^2}{4} > 1 \iff |z| > 2 $$
It should converge for all $z$. Where am I wrong?
 A: In your approach, you seem to consider that $(k!)^\frac{1}{k}$, which should be shown.
Instead, the expression of $\left\lvert a_{k+1} /a_k\right\rvert $ has a nice expression and the limit as $k$ goes to infinity is easy to compute.    
A: [It should converge for all $z$. Where am I wrong?]
I do not know, but you can feel easily that $J(z)$ converges for all $z$, using a majoring series, in fact, for all radius $R>0$, on the disk $D=D(0,R)$, one has  
$$
||\sum^{N}_{k=0}\frac{(-1)^kz^{2k}}{(k!)^22^{2k}}||_D\leq \sum^{N}_{k=0}\frac{R^{2k}}{(k!)}\leq e^{R^2}
$$
and finish easily the reasoning ...
A: Since the exponential function is an entire function, any $f(z)=\sum_{n\geq 0}c_n z^n$ such that $|c_n|\leq\frac{1}{n!}$ is an entire function too. Bessel functions of the first kind arise from the Fourier transform of a compact-supported function, hence they turn out to be entire functions from the Paley-Wiener theorem, too. You computation of the radius of convergence is wrong since the elementary inequality $k!\geq \frac{k^k}{e^k}$ ensures $(k!)^{1/k}\to \color{red}{+\infty}$.
