The Taylor series expansion of $(1+t^2)^{-1}$ is given by
$ 1-t^{2}+t^{4}-t^{6}+\ldots $
A hint in Section 1.1, Exercise 9 in Asymptotic Analysis by Murray gives a remainder term for a truncated Taylor series
$ 1-t^{2}+t^{4}-t^{6}+\ldots+(-1)^{n-1}t^{2n-2} + \frac{(-1)^{n}t^{2n}}{1+t^{2}} $
My first instinct was to look at identities for geometric series but could not get anything in a form that resembled $\frac{(-1)^{n}t^{2n}}{1+t^{2}}$
This results from a high-school identity (which also yields the formula for the partial sums of the geometric series): $$1-u^n=(1-u)(1+u+u^2+\dots+u^{n-1}),\enspace\text{whence}\quad\frac1{1-u}=1+u+u^2+\dots+u^{n-1}+\frac{u^n}{1-u}.$$ Now set $u=-t^2$ to obtain instantly: $$\frac1{1+t^2}=1-t^2+t^4+\dots+(-1)^{n-1}t^{2(n-1)}+\frac{(-1)^nt^{2n}}{1+t^2}$$