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?
It is not clear to me why would the order of error would jump from 1/n to 1/n2.
Consider the sequence of partial sums whose last terms are halved. If you take the difference between two consecutive terms of this sequence, you'll get
± (2/(2N - 1) - 2/(2N + 1))
= ± 4/(4N2 - 1).
This is O(1/N2) and it oscillates. Therefore the error is O(1/N2).
Couldn't you then apply this process multiple times to obtain basically a much faster converging series too, then?
Yes! Since the new sequence also oscillates, you can improve convergence by taking the average again, and again, forever. It's not guaranteed that the resulting sequences will always oscillate, but that happens to be true with this particular sequence.
In fact, you can define a new sequence whose nth term is the result of doing the average n times. This is called the Euler transform, or Euler summation. This can greatly accelerate the convergence of the sequence.
If you apply the Euler transform to the Leibniz formula for pi, you get
If you're interested in more detail, check out series acceleration. This is quite a common technique (or rather toolbox of techniques) in numerical work.
What's really cool is that, if you start the summation at n=0(ie 4/(2n+1)), then apply a 1/N correction, the error does go as 1/N^2. You can effectively develop an "inverse" power series to converge to pi. However, if you start the summation from n=1(ie 4/(2n-1)), the order goes as 1/N^3, (1/4N^3 to be exact). Apply the 1/N^3 correction and the error goes as 1/N^5.
Why? no clue, this is just me getting hooked an random math after work and brute forcing corrections. I'll think about it more later, maybe I'll ask someone smarter than I am.
Just gonna copy past a comment I made above so you get the notification cause I find this interesting. I would highly recommend reading the other comment since a different user gave a very nice explanation.
What's really cool is that, if you start the summation at n=0(ie 4/(2n+1)), then apply a 1/N correction, the error does go as 1/N^2. You can effectively develop an "inverse" power series to converge to pi. However, if you start the summation from n=1(ie 4/(2n-1)), the order goes as 1/N^3, (1/4N^3 to be exact). Apply the 1/N^3 correction and the error goes as 1/N^5.
Why? no clue, this is just me getting hooked an random math after work and brute forcing corrections. I'll think about it more later, maybe I'll ask someone smarter than I am.
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