It is well-known that $te^{-t}$ tends to $0$ as $t$ tends to infinity. But I want to know the decay rate of $te^{-t}$ as $t$ tends to infinity. Using Taylor expansion of $e^{t}$ we have: $${t /e^{t}}=t/(1+t+t^2/2+O(t^3))=1/(t^{-1}+1+t/2+O(t^2))$$
Is the decay rate $t^{-1}$?
In short: no, it is not. Using the Taylor expansions at 0 means that the terms you are neglecting (the $O(t^2)$) are only negligible in a neighborhood of $0$. In particular, when $t\to\infty$, they actually may — and will —dominate...
Now, if you really want to use Taylor series at $0$, the usual trick is to make appear a parameter that tends to $0$ when $t\to \infty$, typically $u=\frac{1}{t}$. In this case, you may want to find the Taylor series (as $u\to 0^+$) of $\frac{1}{u}e^{-\frac{1}{u}}$. Sadly, this will not work in this case (i.e., there is no nice Taylor expansion for this quantity).
The reason, actually, is simple: you want a polynomial approximation of finite order of $t e^{-t}$ when $t\to\infty$, or even one with rational fractions, say. But when $t\to\infty$, the quantity $t e^{-t}$ decays to zero faster than any polynomial or inverse polynomial function in $t$...