Computing 1000 digits of pi using 1648 terms of Newton's infinite series

$MinPrecision = 1000

Pi/24 - Sqrt[3]/32//N

NewtonSum := For[
theSum = 1/12 ; i = 0; term = 1/4//N,
True,
i++,
term /= -4;
term *=  (1/2-i)/(i + 1);
theSum += term/(2 i + 5);
If [Abs[term/(3*(2 i + 7))] < 0.1^($MinPrecision+1), Print[i+1];Return[theSum//N]]
]

NewtonSum

newtonPi = 24*NewtonSum + 3*Sqrt[3]/4

Pi//N