I'm asked to approximate $\pi$ with an error of $10^{-3}$ using that $\tan(\pi /4) = 1$ and the Maclaurin series of $\arctan$ that converges for $-1 < x \leq 1$.
$$ \arctan(x) = \sum_{i=1}^\infty (-1)^{i+1} \frac{x^{2i-1}}{2i-1} $$
For the Lagrange form of the remainder, we have that if $\vert f^{(n+1)}(x) \vert \leq M$ in the interval $[0,1]$ then
$$R_n(1) \leq \left\vert \frac{M \; x^{n+1}}{(n+1)!} \right\vert =\left\vert \frac{M }{(n+1)!} \right\vert $$
Where $R_n(x)$ is the remainder or the error. But I have problems findings this $M$, every time I derivate $f(x)=\arctan(x)$ the values of $f^{(n)}(x)$ at the interval $[0,1]$ get bigger and I can't find a general expression for $f^{(n)}(x)$ . Is there any way to find this $M_n$ such that $ \vert f^{(n)}(x)\vert \leq M_n $ ?
I find out using the computer that for an error of $10^{-m}$ I need $10^m$ terms of the sum, there must be a relation in some part.
You can use the alternating series theorem to bound the error. The error is of the sign of the first neglected term and smaller than it. Since the $501^{st}$ term is $+\frac 1{1001}$ you are there.