This is the (divergent) asymptotic development for $ f(x)=\int_x^{\infty} e^{-{1\over 2}t^2} \ dt $ given by
$$f(x) \sim e^{-{x^2\over 2}}\ (\ {1\over x}\ + \ \Sigma_{k=1}^{\infty}\ {(-1)^k(2k-1)!\over 2^{k-1}(k-1)!}\ {1\over x^{2k+1}}\ ) $$ or $$f(x) \sim e^{-{x^2\over 2}}\ (\ {1\over x}\ - {1\over x^3} + {3\over x^5} - {15\over x^7} + {105\over x^9} - {945\over x^{11}}...)$$
It is easily obtained from an integration by parts. The remainder is given by an explicit integral. From its expression, one can check that $f(x)$ is in fact squeezed between two consecutive sums of the series. As a result, we have the bound, for all $x>0$,
$$0 \geq f(x) - e^{-{x^2\over 2}}\ (\ {1\over x}\ - {1\over x^3} + {3\over x^5} - {15\over x^7} + {105\over x^9}) \geq - e^{-{x^2\over 2}}\ {945\over x^{11}}$$
Divergent series were standard tools at the beginning of the XXe century. I can't provide a reference, but the book of Hardy "Divergent series" may be a starting point for a bibliographic search.
The divergent asymptotics was chosen because it is "good" at infinity. When truncated to some power (say 9), it has the correct behavior when $x$ goes to infinity, that is, it goes to zero, just like $f(x)$. So it gives an interesting approximation of $f$ for all sufficiently big value of $x$. This of course is not the case of a development obtained by truncating a converging series in positive powers of $x$.