Hankel's expansion for the first Bessel function I'm trying to make sense of Hankel's expansion for the first Bessel function $J_0$. According to NIST, it is 10.17(i) here. Using their notation, we have $\nu=0.$ Hence the linked formula becomes:
Let
$$a_k = \frac{(-1^k)1^2*3^2* . . . *(2k-1)^2}{8^kk!}$$
and
$$\omega = z - \pi/4.$$
Then
$$J_0(z) \approx \sqrt{\frac{2}{\pi z}}\left(\cos(\omega)\sum_{k=0}^{\infty}{(-1)^k \frac{a_{2k}}{z^{2k}} - \sin(\omega)\sum_{k=0}^{\infty}{(-1)^k \frac{a_{2k+1}}{z^{2k+1}}}}\right)$$
But that makes no sense, as the series on the right both diverge! The problem is that the absolute value of the numerator of $a_k$ is greater than $(k!)^2$. Therefore, the absolute value of $a_k$ is greater than $\frac{k!}{8^k}$; hence for large $k$, it dominates the $z^{(2k+1)}$ in the denominator.
Can anyone rectify the apparent contradiction? (NIST goofed, maybe?)
 A: $$\frac{(2k-1)!!^2}{8^k k!}=\frac{(2k)!^2}{8^k k! (2k)!!^2}=\frac{(2k)!^2}{32^k k!^3}=\frac{k!}{32^k}\binom{2k}{k}^2 $$
behaves like $\frac{k!}{2^k}$, hence you are right, the series involved in your last expression are not analytic functions of $\frac{1}{z}$. The issue here lies in the meaning of $\approx$. For instance, a continuous but not analytic function of $\frac{1}{n}$ also appears in Stirling's approximation
$$ \log n! \approx n\log\frac{n}{e}+\frac{1}{2}\log(\pi n)+\color{red}{\sum_{k\geq 2}\frac{(-1)^k B_k}{k(k-1)n^{k-1}}} $$
So, here it is the solution of the mistery: the notation $\approx$ means that for any $m\geq 2$ we have
$$ \log n! \approx n\log\frac{n}{e}+\frac{1}{2}\log(\pi n)+{\sum_{k=2}^m\frac{(-1)^k B_k}{k(k-1)n^{k-1}}}+O\left(\frac{1}{n^m}\right)\quad\text{as }n\to +\infty $$
but no further uniformity (about the coefficient hidden in the Landau notation $O$) is granted.
On the other hand, $J_0(z)$ is a non-constant entire function, hence the point at infinity of the Riemann sphere has to be an essential singularity of $J_0(z)$, and that apparently pathological behaviour is not pathological at all.
