Re: Incongruence? hmm...

*To*: mathgroup at smc.vnet.net*Subject*: [mg102713] Re: Incongruence? hmm...*From*: Filippo Miatto <miatto at gmail.com>*Date*: Fri, 21 Aug 2009 04:42:15 -0400 (EDT)*References*: <h6j33u$5j4$1@smc.vnet.net> <4A8D74DC.5050805@gmail.com>

Thank you all for contributing, I didn't see that the sums do not converge for n>2, what a mistake... but still, since they alternate sign for n even or odd, couldn't it be that they sort of 'cancel out' giving that x^3 term in the end? if this is the case, how can i check it with mathematica? Filippo 2009/8/20, Szabolcs Horv=E1t <szhorvat at gmail.com>: > On 2009.08.20. 10:56, Filippo Miatto wrote: >> Dear all, >> I'm calculating the sum >> >> Sum[Cos[m x]/m^4, {m, 1, \[Infinity]}] >> >> in two different ways that do not coincide in result. >> If i expand the cosine in power series >> >> ((m x)^(2n) (-1)^n)/((2n)!m^4) >> >> and sum first on m i obtain >> >> ((-1)^n x^(2n) Zeta[4-2n])/(2n)! > > Hello Filippo, > > I believe the result above to be valid only for n=0 and n=1. For oth= er > values of n the series will not be covergent. > >> >> then I have to sum this result on n from 0 to infinity, but Zeta[4-2n] >> is different from 0 only for n=0,1,2 and the result is >> >> \[Pi]^4/90 - (\[Pi]^2 x^2)/12 - x^4/48 >> >> Three terms, one independent on x, with x^2, one with x^4. >> >> however if I perform the sum straightforwardly (specifying that >> 0<x<2pi) the result that Mathematica gives me is >> >> \[Pi]^4/90 - (\[Pi]^2 x^2)/12 + (\[Pi] x^3)/12 - x^4/48 >> >> with the extra term (\[Pi] x^3)/12. Any idea on where it comes from?? >> Thank you in advance, >> Filippo >> > >