Prove that Zeta(1+1/x) - x = Euler Mascheroni constant as x -> infinity?

Question

Check the following question and Pascal's answer:

http://answers.yahoo.com/question/index;_ylt=AljLVo8RBtg.K8Pg4OSLuXHsy6IX?qid=20070620002146AAOyNlu

Pascal has offered an intriquing question about the limit approaching the Euler Mascheroni constant. See wikipedia:

http://en.wikipedia.org/wiki/Euler-Mascheroni_constant

Prove that in the limiting case where x -> infinity, Zeta(1+1/x) - x = Euler Mascheroni constant. The Zeta function is the sum:

Zeta(k) = Σ (1/n^k) where n = 1 to infinity

Answer 1

this is easy if we look at the laurent expansion of the zeta function.

Zeta(s) = 1/(s-1) + gamma_0+(s-1)gamma_1+(s-1)^2gamma_2+....

where the gamma_n are the stieltes (sp?) constants, specifically gamma_0 is the euler-mascheroni constant.

now simply subsitute s = 1+1/x
zeta(1+1/x) = 1/(1+1/x-1) + gamma_0+gamma_1/x+gamma_2/x^2+...+ gamma_n/x^n

zeta(1+1/x) - x = gamma_0 + gamma_1/x + ... + gamma_n/x^n

as the gamma_n are constants as x -> infinity, this clearly tends to gamma_0 which is the euler-mascheroni constant as required.

QED

Answer 2

i assume you advise locate lim_x->infinity of (one million + one million/x)^x. the concern as suggested initially is trivial. that's the tactic I got here up with in the previous jiffy. that's a well known concern so there may well be greater effective strategies someplace else. (one million + one million/x)^x = ((x+one million)^x)/(x^x)). amplify the numerator with the aid of binomial theorem: (the place C(n,r) is "n p.c. ok" function.) (x+one million)^x = C(x,0)x^x + C(x,one million)x^(x-one million) + ... and observe that purely words C(x,0)x^x, ... , C(x,floor(x/2))x^(x-floor(x/2) have a nonzero x^x coefficient. additionally observe that (with the aid of L'well being facility's rule, to illustrate) we can drop any words with an exponent of decrease than x. The decrease then will become: lim_x->infinity of ((x^x)/0! + (x^x)/one million! + ... + (x^x)/(floor(x/2)!))/(x^x) = lim_x->infinity of one million/0! + one million/one million! + ... + one million/(floor(x/2)!) = e with the aid of fact the means sequence illustration of e^n is (x^0)/0! + (x^one million)/one million! + (x^2)/2! + ... click appropriate answer if useful. thank you!