upper bound of some weird function with an exponencial well I'm trying to find the proof of the following statement, but I cant go forward anymore:

Let $\Omega= \mathbb{R}^{n} \times (0,\infty) $
$$w(x,t)= \sum^{\infty}_{k=0} \frac{1}{(2k)!}\frac{d^kg(t)}{dt^k} ||x||^{2k} $$

where: $$g(t)=e^{-1/t^2}$$

So I need to show that there is a one point $(x^*,t^*) \in \Omega$ such that:
$$w(x^*,t^*) &gt; Ae^{a||x||^2} \space \forall a,A&gt;0$$

so here is what I have:

$$w(x,\frac{1}{\sqrt{t}}) =  \sum^{\infty}_{k=0} \frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!} $$

Hence I wish to study the following difference:
$$D_k(t)=\frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!}-\frac{Aa^k||x||^{2k}}{k!}$$

if there is a $t^*\in (0,\infty)$ such that:
$$D_k(t^*)&gt;0 \space \forall k\in \mathbb{N}$$
then my problem will be over, but there isn't such a $t^*$. So I think I'm in the wrong way but I cannot find another. Thanks for your help.

(oh by the way this function is a solution of the heat equation but i don't think is relevant for this xD)
There inst such a $(x,t) \in \Omega$ since:
$w(x,\frac{1}{\sqrt{t}}) =  \sum^{\infty}_{k=0} \frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!} = e^{-t} cos(||x||) $
There inst such a $(x,t) \in \Omega$ since:
$w(x,\frac{1}{\sqrt{t}}) =  \sum^{\infty}_{k=0} \frac{(-1)^ke^{-t} ||x||^{2k}}{(2k)!} = e^{-t} cos(||x||) $