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

[Werner]

(12-08-2022 06:55 PM)Albert Chan Wrote:  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.

It is far worse than that, I believe. Unless I made a gross error - which is increasingly likely:
Let's try and compute Sk = 1/fk + 1/fk+1 + ... for large k

take k=2^(128-1) so b=128 and k=1.7e38

Sk = ln2*(1/fb + .. 1/fk-1 + 1/fk + 1/fk+1 + ..)
Sk*(1-ln2) = ln2*(1/fb + .. + 1/fk-1)

b=128 = 2^(8-1) so c=8
Sk*(1-ln2) = ln2*ln2*(1/fc + 1/fb-1)

and finally

Sk*(1-ln2) = ln2*ln2*(g4/f4 + g5/f5 + g6/f6 + g7/f7)

for k=2^(128-1), Sk =2.1e-03
for l=2^k-1, Sl = ln2*Sk = 1.5e-03
etc.
so even if the googol^large power term is just below 1e-11, the remainder still adds up to a sizeable fraction..

Cheers, Werner