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

[Albert Chan]

Knowing s ≈ 2.086377665, lets try something easily calculated

sum(1/f, k=16 .. inf) = s - sum(1/f, k=1 .. 15) ≈ 0.262900
sum(1/f, k=2^16 .. inf) ≈ sum(g/f, k=16 .. inf) ≈ 0.262900 * ln2      (*)

Continued doing this, how many (* ln2) we need to make RHS ≈ 1E-11 (1 ULP) ?

number of ln2's = ln(1E-11/0.262900) / ln(ln2) ≈ 65.46

We will need tetration notation to handle term size this enormous!

16 = 2^2^2 = ³2
2^16 = 2^2^2^2 = ⁴2

sum(1/f) terms need for 12-digits accuracy (error < 1ULP) ≈ ⁶⁹2      
I have no words to describe how big this is ...

(*) the estimate had an off-by-1 error. To make it all into equality:

sum(1/f, k=2^16 .. inf) = sum(g/f, k=17 .. inf) = (sum(1/f, k=16 .. inf) - 1/f(16)) * K
where ln2 < K < g(17) ≈ ln2 + 1/2^18

We over-estimated the sum, but under-estimated K.
Overall, we probably over-estimated terms needed.

(0.262900 * ln2) ≈ 0.182228
(0.262900 - 1/480) * (ln2 + 1/2^19) ≈ 0.180785

Udpate: off-by-1 error effect after "first ln2" can be ignored.

number of ln2's = ln(1E-11/0.180785) / ln(ln2) + 1 ≈ 65.44

In terms of tetration exponent, off-by-1 error can be ignored.