I'm trying to gain a better understanding of the difficulties establishing an exact form for $\zeta(2n+1)$

Taking $\zeta(3)$ as a case in point, I can construct the Euler product over primes:

$\displaystyle \zeta(3)=\frac{5}{2}\prod_p\left(\frac{p^7-p^3}{(p^2-1)^2(p^3-1)}\right)$

I can see that when the powers in such products are all even, they behave nicely but in this case (and other odd $s$ the product contains odd powers, making the objective difficult for us. Intuitively for me, this is because the known values for $\zeta(n)$ are closed to even numbers and the values at odd $n$ are in a sense transcendental. Notionally for me, this "closedness" is related to the fact that $\zeta(1)$ diverges so there's perhaps a morphism to some algebra in which $1$ is transcendental.

I have the idea that since $\zeta(2n)$ converges as $n\to\infty$ we need to metrize the distance $\lvert\zeta(2n)-\zeta(n)\rvert$ which goes to zero as $n\to\infty$ and create space not dissimilar to $2-$adics in which we then have something of utility in measuring the distance $\lvert\zeta(n+1)-\zeta(n)\rvert$.

Is this a pre-existing approach and are there any pointers where I might see how this is done?

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    $\begingroup$ I fear that the exact form of $\zeta(3)$, and more generally of $\zeta(2n+1)$ is rather difficult, and the "ideas" here always carry a bit of speculation, which become less and less clear if we proceed with the details. $\endgroup$ – Dietrich Burde Nov 2 '17 at 14:20
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    $\begingroup$ @DietrichBurde thanks for your "comment". Am I right in thinking that since the distance converges to zero, the space created would be complete and have its own calculus? $\endgroup$ – samerivertwice Nov 2 '17 at 14:25
  • $\begingroup$ @DietrichBurde $Q_p$ is complete in the sense that every cauchy sequence converges to a point in $Q_p$ which enables an alternative calculus on $Q_p$. You do realise I was following your lead with the inverted commas!? $\endgroup$ – samerivertwice Nov 2 '17 at 15:34
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    $\begingroup$ Did you look at the methods for evaluating $\zeta(2k)$ ?.. If you want to understand $\mathbb{Q}_p$ then $\zeta(3)$ is clearly not the easiest example. $\endgroup$ – reuns Nov 2 '17 at 21:26
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    $\begingroup$ The methods for evaluating $\zeta(2k)$ in term of the Bernouilli numbers are interesting (real, complex and Fourier analysis). $\mathbb{Q}_p$ is interesting too. But what you write doesn't mean anything and isn't suitable for any mathematical discussion. $\endgroup$ – reuns Nov 3 '17 at 0:06

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