radius of convergence of hypergeometric functions

Question

Hypergeometric function of scalar arguments is defined as \begin{eqnarray} _aF_b\left(p_1,...,p_a;q_1,...,q_b;z\right) &=&\sum_{i=0}^{\infty} \frac{\left(p_1\right)_i...\left(p_a\right)_i}{\left(q_1\right)_i...\left(q_b\right)_i}\frac{z^i}{i!} \end{eqnarray} In general, it is told that if $a\le b+1$, and if $z$ is finite, then the radius of convergence is $\infty$. But I am facing a problem. I am trying to calculate $_1F_1\left(1,2,10^6\right)$, but the series diverges. Such high values of $z$ can occur in stochastic systems, where the noise$\left(\text{variance}\left(\sigma^2\right) = 1e-6\right)$ is very low and $z = \frac{\text{some finite value between 0 and 1}}{\sigma^2}$. One can see that in this case, $z$ is finite and $a\le b$, but the function still diverges. So, is there anything that I am missing here regarding the conditions for the radius of convergence?