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Re: Incongruence? hmm...

  • To: mathgroup at
  • Subject: [mg102713] Re: Incongruence? hmm...
  • From: Filippo Miatto <miatto at>
  • Date: Fri, 21 Aug 2009 04:42:15 -0400 (EDT)
  • References: <h6j33u$5j4$> <>

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?

2009/8/20, Szabolcs Horv=E1t <szhorvat at>:
> 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=
> 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

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