Summing divergent asymptotic series I found the sine integral si to be
$$Si (x)\sim \frac \pi 2+\sum _{n=1}^\infty (-1)^n \left(\frac{(2 n-1)! \sin (x)}{x^{2 n}}+\frac{(2 n-2)! \cos (x)}{x^{2 n-1}}\right)$$
Say I want to find $Si(\frac \pi 4)$ what options have I got to use this divergent series to find the actual value?

In order to find a good approximation of $\text{Si}\left(\frac{\pi}{4}\right)$, I strongly suggest you to use a converging series and not a diverging one. For instance, the almost trivial:
$$\text{Si}\left(\frac{\pi}{4}\right)=\int_{0}^{\pi/4}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!}x^{2n}\,dx = \sum_{n\geq 0}\frac{(-1)^n \pi^{2n+1}}{4^{2n+1}(2n+1)(2n+1)!}$$
converges pretty fast.
