Sorry to bother you, but I guess that this question is not appropriate for MO, so I repost it here hoping that someone could give me a clue.
I would like to have an estimate for the series $$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m,$$ where $e$ is the base of natural logarithm, $k!$ is the factorial of the integer $k$, $t$ represents the time and $t>0$, $m$ is a positive integer and $m>1$ (Obviously, $P(t)=1$ when $m=1$ since $P(t)$ is exactly the cdf of Poisson distribution with associated parameter $t$ in this case.).
I am interested in showing that the order of the scale could look something like $$P(t) = O(t^{-\alpha m}),\alpha > 0.$$
But I find it difficult for me to extend their method to the cases when $m>2$.
I also tried to calculate $P(t)$ in Mathematica and it gave me the result like
$$P(t) = \sum\limits_{k = 0}^\infty (e^{-t}\frac{t^k}{k!})^m = e^{-mt}\text{HypergeometricPFQ}[...,t^m]$$
It seems that the decreasing speed of $P(t)$ is much slower than that of $e^{-mt}$ as $t$ increases. But I could not find the closed-form bound for $P(t)$ when $m>2$ just like $O(1/\sqrt t)$ when $m=2$.
Does anyone know of a scale or a bound of $P(t)$ in the literature? Any comments and answers would be highly appreciated. Many thanks!
The function you have, and this is also the closed form that Mathematica gave you, is related to the hypergeometric function, $$ P(t) = e^{-m t}\,\,{}_0F_{m-1}(;1,1,\ldots,1; t^m). $$ To calculate its behaviour as $t\to\infty$, at least in this case, is relatively easy. In particular the DLMF gives the formulas necessary, 16.11.9 and 16.11.1-4.
The result is $$ P(t) = \frac{1}{(2\pi)^{\frac{m-1}{2}}\sqrt{m}}\frac{1}{t^{\frac{m-1}{2}}} + O(t^{\frac{-1-m}{2}}). $$
You can also check this for small values of $m$ using Mathematica's function
Series, which understands how to calculate asymptotic behaviour.