I gurgled,
Let H := Harmonic(K) and g:= Euler's constant (gamma ~ .577), then
H - g 1 1 K = e - - - --------------- + O(??), 2 H - g 1 24 (e + -) 2
where ?? is < a finite negative power of e^(H-g)+1/2 !
Bullbleep.
(I.e., these three terms satisfy the Euler-Maclaurin equation identically.) This is just an artifact of an irrevertible divergent series. Newton's method works just fine:
H - g 1 1 1 K = e + - - --------------- - ---------------- 2 H - g 1 H - g 1 2 24 (e + -) 48 (e + -) 2 2 11 7 5227 - ------------------ + ------------------ + --------------------- + ... H - g 1 3 H - g 1 4 H - g 1 5 1920 (e + -) 3840 (e + -) 2903040 (e + -) 2 2 2 But since e^H is large, all you need are the first two terms. --rwg Save the DOUBLE-STRAND BLADDERSNOUT.