Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$.

  There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.

Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$. There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.

Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$.

There are also nice formulas for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$.

There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.

Well, I wish you the best. But I don't think you are going to succeed for two reasons:

(1) Proving almost anything about $\zeta(2k+1)$ is hard.

(2) There are nice formulas for $\sum_{n=1}^{\infty} \cos (n \theta)/n^{2k}$ and for $\sum_{n=1}^{\infty} \sin (n \theta)/n^{2k+1}$.

There are not particularly nice formulas for $\sin$ with even powers or $\cos$ with odd. One way to think about this is that the imaginary part of $\log (1-e^{i \theta})$ has a simple formula but the real part does not.