I know the formula:
$\arctan(x) = \frac{\pi}{2} - \arctan\left(\frac{1}{x}\right)$
and it's easy to find the appropriate Taylor expansion of $\arctan\left(\frac{1}{x}\right)$, but my problem is in finding upper bound (and such number $n$) of Lagrange Remainder, and proving that this remainder is less than $0.001$
$$\arctan(4)=\frac{\pi}{2}-\sum_{n\geq 0}\frac{(-1)^n}{4^{2n+1} (2n+1)}\tag{1}$$ and by Leibniz' test $$\left|\sum_{n\geq 2}\frac{(-1)^n}{4^{2n+1}(2n+1)}\right|\leq \frac{1}{4^5\cdot 5}\tag{2}$$ hence a good approximation (i.e. with an error within $10^{-3}$) is given by $$ \arctan(4)\approx\frac{\pi}{2}-\sum_{n=0}^{2}\frac{(-1)^n}{4^{2n+1}(2n+1)}=\frac{\pi}{2}-\frac{3763}{15360}=\color{green}{1.325}8\ldots\tag{3}$$