Why isn't there a formula for $\zeta(k)=\sum_{n=1}^\infty\frac{1}{n^k}$ involving $\pi$ when $k$ is odd? Now we know that $\sum \frac{1}{n}=\text{divergent}, \sum \frac{1}{n^2}=\frac{\pi^2}{6}$ but now this for $\sum \frac{1}{n^3}=1.20....$ and again $\sum \frac{1}{n^4}=\frac{\pi^2}{90}$ .Now somewhere there is a general term when $n$ has power $2k$ where $k$ is a positive integer. Why doesn't there exist a $\pi$ series for odd numbers like $n^3$, $n^5$?

In Euler's solution to the Basel problem, he factors sine into an infinite product of binomials, 
$$ \frac{\sin( z)}{z} =\prod_{n=1}^\infty \left(1-\frac{z^2}{\pi^2 n^2}\right) , $$
these binomials have the nice form of a difference of squares because the roots of sine are symmetric about the origin. The roots are also spaced by integer units of $\pi$ which is the key to solving the problem. 
Because sine's roots are symmetric, we can rotate the argument of sine in the complex plane to get a difference of $4$'th powers or a difference of $6$'th powers and so on. For instance we can write, 
$$ \frac{\sin(z)}{z}\frac{\sin(iz)}{iz} =\prod_{n=1}^\infty \left(1-\frac{z^4}{\pi^4 n^4}\right) , $$
this essentially follows from the identity $(1-(iz)^2)(1-z^2)=(1-z^4)$. 
If we could somehow get a difference of cubes we could solve $\sum 1/n^3$ using, 
$$ \text{(Some function with a known Taylor series)}=\prod_{n=1}^\infty \left(1-\frac{z^3}{n^3} \right).$$
The problem is that the cubed roots of unity are not symmetric about the origin. Because of this we can't use sine.
We could use the reciprocal of the Gamma function, which has roots only at the negative integers,
$$ \frac{1}{\Gamma(-z)}\frac{1}{\Gamma(-e^{i2\pi/3}z)} \frac{1}{\Gamma(-e^{-2i\pi/3}z)} \ \propto \  \prod_{n=1}^\infty \left(1-\frac{z^3}{n^3} \right),$$
but unfortunately the relevant Taylor coefficient is only known in terms of $\zeta(3)$ which is what we would like to discover. 
