The original function is $$ f(x)=e^{-x^2}$$ finding the series expression of that function results in
$$\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{n!} $$ Then integrating that series expression we get
$$\sum_{n=0}^\infty \frac{-x^{2n+2}}{(n+1)!}$$
So if I want to find when the value of the above Maclaurin sum when it is less than $0.001$ I just keep increasing the $n^{th}$ terms till I get an answer which i found to be when $n=9$ and the value $0.0000311$. Is this the correct way of solving it or is there a easier way?