>And if you need to "produce" pi just remember pi/4 = 1 - 1/3 + 1/5 - 1/7...
This converges too slowly to be practically useful.
If you group the consecutive terms in pairs you get that the nth pair sums to 1/(2n + 1) - 1/(2n + 3) = (2n+3 - (2n+1))/(2n+1)(2n+3) = 2/(2n+1)(2n+3) = Theta(1/n^2). Thus it has the same asymptotic growth order as sum 1/n^2. That has has monotone terms and so it's easy to estimate how many terms we need to get k correct fractional digits by solving 10^(-k) = 1/n^2, giving n = 10^(k/2).
This is off by a big constant factor but it gives you the right idea that you need an exponential number of terms relative to the desired number of significant digits. From Wikipedia: "After 500,000 terms, it produces only five correct decimal digits of pi."
So, this series for pi has only theoretical relevance.
This converges too slowly to be practically useful.
If you group the consecutive terms in pairs you get that the nth pair sums to 1/(2n + 1) - 1/(2n + 3) = (2n+3 - (2n+1))/(2n+1)(2n+3) = 2/(2n+1)(2n+3) = Theta(1/n^2). Thus it has the same asymptotic growth order as sum 1/n^2. That has has monotone terms and so it's easy to estimate how many terms we need to get k correct fractional digits by solving 10^(-k) = 1/n^2, giving n = 10^(k/2).
This is off by a big constant factor but it gives you the right idea that you need an exponential number of terms relative to the desired number of significant digits. From Wikipedia: "After 500,000 terms, it produces only five correct decimal digits of pi."
So, this series for pi has only theoretical relevance.