I know that for infinite series and $|z|<1$ there exists a confluent hypergeometric expression 


$
\sum\_{k=0}^{\infty} \frac{z^k}{k!k!} = F\_{1}[;1;z]
$


This is not very helpful though, and I 'd like to know if it is possible to get some asymptotic expansion for this function and if there exists some general approach to bounding hypergeometric functions asymptotically. 



Asymptotic bounds for a confluent hypergeometric function $F_{1}[;1;x]$
This particular function can be expressed in terms of Bessel's I function as $I(0,2\sqrt{z})$, and from there an asymptotic expression (at $\infty$) is easily derived. It starts
$$\frac{e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{1}{4}\right)}}{2\sqrt{\pi}} + O\left(e^{\frac{2}{\sqrt{\frac{1}{z}}}}\left(\frac{1}{z}\right)^{\left(\frac{3}{4}\right)}\right)$$
where the roots are chosen to have the correct branching behaviour.


The easiest way to obtain such results is actually from the ODE satisfied by your function, in this case $zy'' + y' - y$ with $y(0)=1$. It is easy to get an ODE at $\infty$ from there, and from there one gets the asymptotic expansion. The hardest part is getting the singular behaviour 'just right', as well as the branches. Those who have voted to close this likely have never had to compute the asymptotic expansion along a branch cut with the expansion point being an irregular singularity. While it is known how to do this, it is practiced by very very few.


