Find Discrete Time Fourier coefficients of $(-1)^n x[n]$ Given that $x[n]$ is an N-periodic sequence with Fourier coefficients $a_k$, I want to find the Fourier coefficients of
$$(-1)^n x[n]$$
for the situation in which $N$ is odd.  I'm also interested in the case of when $N$ is even, but the odd scenario is more important at the moment.
I believe the property
$$x[n]y[n] \Rightarrow b_k = \sum_{m=0}^{N-1} a_m b_{k-m}$$
would be useful.  I know that this also matches the form of periodic convolution.  I have up to this point determined the Fourier coefficients of the $(-1)^n$ portion.  The Fourier coefficients are given by
$$(a_k)_{k=0}^{N-1} = \frac{1}{N} \sum_{n=0}^{N-1} x[n] \cdot \mathrm{e}^{-\mathrm{j}2 \pi n k /N}$$
which for $(-1)^n$ are given by
$$a_k = \frac{1}{N} \frac{1 - \mathrm{e}^{\mathrm{j} \pi (N-2k)}}{1-\mathrm{e}^{\mathrm{j} \pi (1-\frac{2k}{N})}}$$
where we used a geometric series to eliminate the sum.  I'm kind of lost as to where to proceed from here.  I feel like there's probably a really straight forward answer to this question that I'm missing and things start getting really complicated when I attempt to plug into the property.  Any suggestions?
 A: Let's make notation clear first. For general $x \in \mathbb{R}^N$, define
$$
\hat{x}_k = \frac{1}{N} \sum_{l=0}^{N-1}x_n\exp(-i2\pi lk/N)
$$
Consider an arbitrary $x \in \mathbb{R}^N$, and let $y \in \mathbb{R}^N$ be defined such that $y_n = (-1)^n$. Let $z \in \mathbb{R}^N$ be defined such that $z_n = x_ny_n$. We're looking for $\hat{z}$. The convolution theorem says:
$$
\hat{z}_k = (\hat{x} * \hat{y})_k = \sum_{l=0}^{N-1} \hat{x}_{k-l} \; \hat{y}_{l}
$$
where the convolution is periodic. So first we have
$$
\hat{y}_k = \frac{1}{N} \sum_{l=0}^{N-1} \exp(i\pi l(1-2k/N))
$$
Note that if $N$ is even, that $k=N/2 \Rightarrow \hat{y}_{N/2} = 1$. When $k\neq N/2$, the geometric sum formula you stated holds so that:
$$
\hat{y}_k = \frac{1}{N} \frac{1-\exp(i\pi(N-2k))}{1-\exp(i\pi(1-2k/N)}
$$
In this case, notice that the numerator is zero always when $N$ is even (excluding when $k=N/2$). Thus, when $N$ is even,
$$
\hat{y}_k = \delta[k-N/2]
$$
so that
$$
\hat{z}_k = \hat{x}_{k-N/2}
$$
I'm not so sure that things work out so nicely in the $N$ odd case, but maybe you can give it a try.
