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

[Albert Chan]

I was curious what it take if we do brute force, no tricks, S = sum(1/F)

G(b)/F(b) = sum(1/F(k), k = 2^(b-1) .. 2^b - 1)

S ≈ 2.086, for 12-digits accuracy we solve G(b)/F(b) = 1E-11 (1 ULP), for b

G(b)/F(b) ≈ ln(2)/(b*log2(b)) ≈ 1E-11 --> b ≈ 2.2E9      (*)

It take more terms for convergence, even though all other terms below 1 ULP.

sum(1/F) terms needed > 2^b ≈ googol ^ 6622660

(*)
We assume F grow about same rate as primes, to be conservative.
We know F grow faster, because sum of reciprocal primes diverges.

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Update: using actual recursive definition of f(b)

ln(2)/f(b) ≈ 1E-11 --> b ≈ 85573726

sum(1/F) terms needed > 2^b ≈ googol ^ 257603

Edit: google should be googol (10^100), corrected.