Yes, the field is infinite, but it is only log divergent near the plate, so that it is hard to see the divergence numerically. You can see this easily by solving the problem of a uniformly charged infinite plate, which is a 2d problem. Here the charges are uniform along the negative real axis, where the 2d space is imagined to be the complex plane.
This problem can be understood as follows: the 2d electrostatic field of a point charge at the origin, written as a map from C to C can be written in complex form as:
$$ E_x + i E_y = \frac{z} { 2\pi |z|^2 }= \frac{1}{ 2\pi \bar{z}}$$
it points radially outwards. This is a pure antiholomorphic function, except at the origin. It's more familiar to deal with holomorphic functions, so conjugate it!
$$ E_x - iE_y = E(z) = {1\over 2\pi z} $$
Now you want to superpose all the charges on the negative z axis. This is a simple integral:
$$ \int_{-\infty}^0 E(z-a) da = - {1\over 2\pi} \log(z) $$
where I threw away an infinite additive constant (you should think of this as calculating the potential difference between the point z=1 and any other point). This is the function with a given fixed cut discontinuity on the negative real axis.
So at the point $r,\theta$, the electric field is
$$ E_x = - {\rho\over 2\pi} \log(r) $$
$$ E_y = {\rho \theta\over 2\pi} $$
Where I have restored the $\rho$. The part in the y-direction is finite, as your intuition says--- the discontinuity is equal to the charge density (this charge density is the cut discontinuity of the electric field analytic function, which is a way of making it obvious that the electric field goes as the log--- the log function as a constant cut discontinuity). The divergent part is in the x direction, and it is only invisible in the bulk disk because when you get close to the surface, you have cancellations from the left and from the right that wash it out.
So the answer is yes, the E field is log divergent, but only the component in the plane of the disk pointing out. The solution of the disk asymptotes to the plane solution in the near disk limit.
If a small piece of your disk is located at
$$
x=r\cos\theta\, ,\qquad y=r\sin\theta,\qquad z=0
$$
then the density would be $dq=\sigma dA=\sigma r dr d\theta$ where $\theta$ is the polar angle on the disk since $dA=r dr d\theta$ is the area of a small piece of your disk, with radius $r$ and arclength $rd\theta$.
For a point located on the $\hat z$ axis at $Z\ne 0$, this small amount of charge will produce the infinitesimal field
$$
d\vec E=\frac{dq}{4\pi\epsilon_0} \frac{Z\hat z- r\cos\theta\hat x - r\sin\theta\hat y}{(r^2+Z^2)^{3/2}}\, .
$$
The full field is obtained by integrating the infinitesimal fields.
Best Answer
$$A=\pi r^2$$ $$\frac{dA}{dr}=\pi\cdot2r$$ $$dA=2\pi rdr$$
Alternatively, you can write : $\lim_{\Delta r\to 0}\frac{\Delta A}{\Delta r}=\lim_{\Delta r\to 0}\frac{\pi\{(r+\Delta r)^2-r^2\}}{\Delta r}=\lim_{\Delta r\to 0}\frac{2\pi r\Delta r+\Delta r^2}{\Delta r}=2\pi r+0$
You have to ignore $(dr)^2$ as it is very small. Why? Because you took the limit while taking infinitesimal rings.