In this paper we will try to find a solution for the Riemann Zeta function for odd integers. We will start with ζ(3) (the Riemann Zeta function with s=3) emulating the "Basel problem". But instead of using a sine or cosine function, using functions similar to these:f(x)=1-x^3/3!+x^6/6!-x^9/9!+⋯f(x)=1/3 (e^(-x·e^(1/3 (2πi) ) ) 〖+e〗^(-x·e^(2/3 (2πi) ) )+e^(-x·e^(3/3 (2πi) ) ) )We will discover that the process itself seems ok, but with a problem. The solutions of the above functions are not periodic, so we cannot emulate the "Basel problem" perfectly, obtaining the following value:ζ(3)=(π^3/(3!(√3)^3 ))/(1-1/2^3 )=1.1366020≠1.202056903With a small correction, we arrive to:ζ(3)=(π^3/(3!(√3)^3 ))/(1-1/2^3 ) e^(2/√3)/3=1.202173775≈1.202056903 [7]But not getting the correct value anyhow. The only way of obtaining the correct value would be to find a function of the form:g(x)=1-r(3)·x^3/3!+r(6)·x^6/6!-r(9)·x^9/9!+⋯That has periodic zeros. Where r(n) is an unknown function to be calculate/discovered.We have also generalized this study to calculate a general ζ(k) where k can be higher odd numbers, or even numbers. Having ζ(k) for even numbers would lead to obtaining a closed equation for the Bernoulli numbers. If a generalization for k as a general complex number was possible, we could even consider k=½+it, obtaining a closed function for the zeros of the Riemann Zeta function.