DanA writes
I've long wondered what is the smallest n for which
(*) 1 + 1/2 + 1/3 + . . . + 1/n > 100.
Roughly, I thought it was where ln(n) = 100, i.e.,
n ~ e^100 ~ 2.688 x 10^43.
But more exactly, it would be closer to where
100 - ln(n) = gamma (Euler's), i.e. let K = floor( e^(100-gamma) ).
Then n should be fairly close to K (which is roughly only 1.509 x 10^43).
Mathematica gives K as exactly
K = 15092688622113788323693563264538101449859497
and to my surprise it claims that
1 + 1/2 + 1/3 + . . . + 1/K > 100 but 1 + 1/2 + 1/3 + . . . + 1/(K-1) < 100 (!).
Can someone confirm that this K is exactly the least n for which (*) holds?
Thanks,
Dan
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 ! (I.e., these three terms satisfy the Euler-Maclaurin equation identically.) E.g., For H = 100, this instantly gives K ~ 15092688622113788323693563264538101449859496.864101
P.S. Anyone know a useful upper bound U(n) for the error
E(n) = |(1+1/2+1/3+...+1/n) - ln(n) - gamma| ? Evidently E(n) gets very small very fast.
Use the Euler-Maclaurin expansion of H(n) = Digamma(n+1) + gamma. The lead term is ~ 1/(2n+2). You have a big n. --rwg SOUP-KITCHEN UP TO HIS NECK PS, MIT AIMemo 304, (Acceleration of Series) (if anyone can find one) transforms H(n) into another n term series which "converges faster", i.e., the lead terms are >> the final ones.