radius of convergence of hypergeometric function

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Looking up information on the Bessel function there is a formula as $|z| \to \infty$:

$$ I_0(z) \approx \frac{e^z}{\sqrt{2\pi z}} {}_2F_0( \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2z}) = \frac{e^z}{\sqrt{2\pi z}} \left[ \sum_{k=0}^n \frac{(\tfrac{1}{2})_k(\tfrac{1}{2})_k}{k!}\frac{1}{(2z)^k} \right] $$

The infinite series on the right doesn't seem to converge anywhere, it's radius of convergence is $0$. Is it possible to recover a value using borel summation?


COMMENT The series I have written above terminates and is therefore wrong. Either it could read:

$$ I_0(z) \approx \frac{e^z}{\sqrt{2\pi z}} {}_2F_0( \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2z}) = \frac{e^z}{\sqrt{2\pi z}} \left[ \sum_{k=0}^\color{red}{\mathbf{n}} \frac{(\tfrac{1}{2})_k(\tfrac{1}{2})_k}{k!}\frac{1}{(2z)^k} + O\left( \frac{1}{z^{n+1}} \right) \right] $$

If we try to complete the infinite series, I believe the radius of convergence is $0$, which is why I ask about Borel summation.

$$ I_0(z) \approx \frac{e^z}{\sqrt{2\pi z}} {}_2F_0( \tfrac{1}{2}, \tfrac{1}{2}, \tfrac{1}{2z}) = \frac{e^z}{\sqrt{2\pi z}} \left[ \sum_{k=0}^\color{gray}{\mathbf{\infty}} \frac{(\tfrac{1}{2})_k(\tfrac{1}{2})_k}{k!}\frac{1}{(2z)^k} \right] $$

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The $n$ in the summation index should be an $\infty$, and the series is meant to be interpreted as an asymptotic expansion, where convergence is not a consideration.