# Using Taylor(Maclaurin) expansions at infinity

To the best of my knowledge, we usually say that the expansion is most accurate around the expansion point (say, around $$x=0$$ in the case of the Maclaurin series).

I know the question is most likely stupid, but I often notice that the Taylor series is often used when working with infinite series where $$n \to \infty$$ and we are not in the neighborhood of the expansion point. What is the formal explanation to why we can safely do this without worrying about the accuracy etc. ?

## 1 Answer

We can only use a Taylor series to approximate a function within its "radius of convergence", which is the values of $$x$$ where the Taylor series actually approximates the function. So, we could only use the Taylor series at a "neighborhood of the expansion point", as you say, if it's within the radius of convergence.

This is why you can use the Taylor series for $$\sin x$$ and $$e^x$$ for any $$x$$ (since their radii of convergence include all of $$\mathbb{R}$$) but not for the series expansion of $$\frac{1}{x^2+1}$$, whose radius of convergence only includes $$(-1,1)$$.