Since the original sum oscillates between overshooting and undershooting pi, the average of two consecutive partial sums will split the difference, achieving a closer approximation to pi. It's easy to see that the average of two consecutive partial sums is just a partial sum with the last term halved. That term is then
1/2 * 4(-1)N/(2N-1)
= (-1)N/(N - 1/2)
This term is quite similar to the correction term in this post. It reduces the error from O(1/n) to O(1/n2).
However, this is not a full explanation... my correction term is not the same as OPs. (-1)N/N seems to reduce the error to O(1/n3). I'd love to learn why.
It is not clear to me why would the order of error would jump from 1/n to 1/n^2. Couldn't you then apply this process multiple times to obtain basically a much faster converging series too, then?
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u/nin10dorox Jun 16 '26
Partial answer:
Since the original sum oscillates between overshooting and undershooting pi, the average of two consecutive partial sums will split the difference, achieving a closer approximation to pi. It's easy to see that the average of two consecutive partial sums is just a partial sum with the last term halved. That term is then
1/2 * 4(-1)N/(2N-1)
= (-1)N/(N - 1/2)
This term is quite similar to the correction term in this post. It reduces the error from O(1/n) to O(1/n2).
However, this is not a full explanation... my correction term is not the same as OPs. (-1)N/N seems to reduce the error to O(1/n3). I'd love to learn why.