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Resummation of the Harmonic series!

by yasiru89
Tags: harmonic, resummation, series
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yasiru89
#1
Oct26-08, 02:12 AM
P: 107
We have different summation senses under which Cauchy-divergent series can be summed to finite values. I was wondering if such a procedure existed for the Harmonic series, [itex]\sum_{n = 1}^{\infty} n^{-1}[/tex].

I'm putting this in the number theory discussion since the obvious connection with the Riemann zeta-function's pole at unity. However this guarantees there's no basic zeta regularization to the harmonic- so is there a deeper zeta-based result? Or some other way in which a value might be assigned to it? (the only one I know of is the Ramanujan sum of [itex]\gamma[/tex], which is just based on the definition of Euler's constant in the case of the harmonic- and expressing this in terms of other constants has so far proven futile)
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mhill
#2
Oct26-08, 03:01 AM
P: 193
of course Yasiru since [tex] \zeta (1) [/tex] is infinite the regularization procedure is useless, this is a pain in the neck but can be solved via ramanujan summation

[tex] S = \sum_{n=1}^{N}a(n)- \int_{1}^{N} dx a(x) [/tex]

and taking N--->oo if you set a(n)=1/n (Harmonic series) you would get

[tex] \sum_{n=1}^{N}1/n = \gamma [/tex] (Euler's constant)
yasiru89
#3
Oct26-08, 06:58 AM
P: 107
I wonder though- is that the only one that works for the harmonic? Its not as impressive as it could be since we end up with the definition of Euler's constant as the constant of the series in the Euler-Maclaurin sum formula.


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