----- Original Message ----- From: "R. William Gosper" <rwg@osots.com> To: <math-fun@mailman.xmission.com> Sent: Friday, November 10, 2006 07:14 Subject: [math-fun] Inverse HarmonicNumber oops
I blurted
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
Off by 1:
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
You might prefer the asymptotic series which I gave in "Inverse of Harmonic Numbers", sci.math, 2006 Apr. 4: <http://groups.google.com/group/sci.math/msg/0926db8773d69b81>. You had also said in your previous post:
But since e^H is large, all you need are the first two terms.
Although I certainly suspect this is true (assuming we know that our result must be an integer), I know of no proof, only a heuristic argument. David W. Cantrell