\frac{\pi}{4} = \sum_{n=0}^{\infty} \left(\frac{1}{4n+1}-\frac{1}{4n+3}\right) = \sum_{n=0}^{\infty} \frac{2}{(4n+1)(4n+3)} .
which can be evaluated to high precision from a small number of terms using Richardson extrapolation or the Euler-Maclaurin formula.