So a buddy of mine and I are sitting in our signals class, and he asks me a question: if the coefficients of a discrete time Fourier series are period, what are their Fourier coefficients? I understand that the Fourier transform of a Fourier transform returns a translated version of the original function, is this true with series as well?
For a DT signal $x[n]$ with period $N$ and frequency $\omega_0 = \frac{2\pi}{N}$, the Fourier coefficient $a_k$ is
$$a_k = \frac{1}{N}\sum_{n=\langle N \rangle} x[n]e^{jnk\omega_0}$$
If we treat $a_k$ as a periodic signal $a[n] = a_n$ (we know it will be periodic with the same period), then its Fourier coefficients $b_k$ are
$$b_k = \frac{1}{N}\sum_{m=\langle N \rangle} a[m]e^{jmk\omega_0} = \frac{1}{N}\sum_{m=\langle N \rangle}\left( \frac{1}{N}\sum_{n=\langle N \rangle} x[n]e^{jnm\omega_0}\right)e^{jmk\omega_0}.$$
Is there any noticeable comparison between $x[n]$ and $b[n] = b_n$?