Compute $\int_{0}^{\frac{1}{2}} \frac{\arctan x - \sin x}{x}\, dx $ with an error less than $10^{-2}$ I have used the Taylor expansions of the functions arctan and sin  and the fact that power series uniformly converge  in $[0,1/2]$ so I obtained: 

$$\sum \biggl(\frac{1}{(2n+1)!}-{\frac{1}{2n+1}}\biggr) \frac{1}{(2n+1)2^{n+1}}$$

How can I estimate it?
 A: Using Taylor expansions$$\frac{\tan ^{-1}(x)-\sin (x)}{x}=\sum_{p=1}^\infty \frac{(-1)^p ((2p)!-1)}{(2 p+1) (2p)!}x^{2p}$$
$$I=\int_0^{\frac 12}\frac{\tan ^{-1}(x)-\sin (x)}{x}\,dx=\sum_{p=1}^\infty \frac{(-1)^p  ((2p)!-1)}{2^{2 p+1}(2 p+1) (2p+1)!}$$ that you can rewrite as
$$I=\sum_{p=1}^n \frac{(-1)^p  ((2p)!-1)}{2^{2 p+1}(2 p+1) (2p+1)!}+\sum_{p=n+1}^\infty \frac{(-1)^p  ((2p)!-1)}{2^{2 p+1}(2 p+1) (2p+1)!}$$
This is an alternating series; so you look for $n$ such that
$$\frac{ (2 n+2)!-1}{2^{2 n+3} (2 n+3)^2  (2 n+2)!} \lt \epsilon$$ Let us neglect the $1$ in numerator and this reduces to
$$\frac{1}{2^{2 n+3}(2 n+3)^2}\lt \epsilon\implies {2^{2 n+3}(2 n+3)^2}\gt \frac 1\epsilon$$
The sequence defined by $a_n=2^{2 n+3}(2 n+3)^2$ is (easy to compute with your phone)
$$\{72,800,6272,41472,247808,1384448,7372800\}$$ So, one term is not enough; what about two terms ?
If I am not mistaken, this would give $-\frac{331}{57600}\approx -0.005747$ while numerical integration would give $-0.005885$
