[VA] SRC #012c - Then and Now: Sum

[Albert Chan]

Sorry about math sessions. I will keep it short.

Let F(n) = n * F(bits of n), except that F(2)=2, F(1)=1
Let G(n) = sum of n-bits integer reciprocal, except that G(1)=5/4

Let sum = index from 1 to K-1, SUM = index K to infinity

S = sum(1/F) + SUM(1/F)       // definition
S = sum(G/F) + SUM(G/F)      // sum(G/F) accelerated convergence, but still not fast enough

S = sum(G/F) + LN2 * SUM(G/LN2/F)

Since G/LN2 ≥ 1, no matter how big K is, we have:

S ≥ sum(G/F) + LN2 * SUM(1/F)
S ≥ sum(G/F) + LN2 * (S - sum(1/F))

S*(1-LN2) ≥ sum((G-LN2)/F)

No need to hard code conditions for index K, or for G converged to LN2.
Just sum RHS until convergence. It will converge, very quickly.

(G-LN2) part shrink at O(1/2^n), which already can converge without F