
#1
Mar2707, 03:20 PM

P: 76

if [tex] g(s)= \sum_{n=1}^{\infty} a(n) n^{s} [/tex]
Where g(s) has a single pole at s=1 with residue C, then my question/conjecture is if for s >0 (real part of s bigger than 0) we can write [tex] g(s)= C(\frac{1}{s1}+1)s\int_{0}^{\infty}dx(CxA(x))x^{s1} [/tex] of course [tex] A(x)=\sum_{n \le x}a(n) [/tex] the question is if the series converge for s >1 with a pole there is a method to 'substract' this singularity (pole) at s=1 to give meaning for the series at any positive s. I think that the 'Ramanujan resummation' may help to give the result: [tex] \sum_{ n >1}^{[R]}a(n)n^{s} = g(s)C(s1)^{1} [/tex] valid even for s=1 or s>0 (??) 



#2
Mar2707, 04:39 PM

P: 461

Jose, how's it been?



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