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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 $0.001$

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$$\arctan(4)=\frac{\pi}{2}-\sum_{n\geq 0}\frac{(-1)^n}{4^{2n+1} (2n+1)}\tag{1}$$ 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 $$ \arctan(4)\approx\frac{\pi}{2}-\sum_{n=0}^{2}\frac{(-1)^n}{4^{2n+1}(2n+1)}=\frac{\pi}{2}-\frac{3763}{15360}=\color{green}{1.325}8\ldots\tag{3}$$

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    Thank you for your answer! My problem is that I need to proof this error only using the formula for Lagrange Remainder, taking $n$-th derivative of $\arctan$ (I'm not sure if it is even possible, but i didn't study Leibniz' test, and in this problem I'm talking about finite sums) – 2017-02-16
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    @puhsu: well, in such a case you have to replace the (quite trivial) $(2)$ with a bound for the fifth derivative of the arctangent function in a neighbourhood of the origin, that is a bound for the fourth derivative of $\frac{1}{1+x^2} = \frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right)$. – 2017-02-16
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    That is not that painful: the function $\frac{24 \left(1-10 x^2+5 x^4\right)}{\left(1+x^2\right)^5}$ attains its maximum at the origin. – 2017-02-16