Taking $x[n] = cos({\pi}n)$ and $h[n] = \left(\frac{-1}{2}\right)^{(n-2)}$ is the convolution sum $y[n] =x[n] \ast h[n] = 0$? Consider an LTI system with signal input $x[n] = cos({\pi}n)$ and impulse response $h[n] = \left(\frac{-1}{2}\right)^{(n-2)}$ what is the convolution sum $y[n] =x[n] \ast h[n]$?
My answer for now:
Since the convolution is 
$$y[n] = \sum_{k=-\infty}^\infty x[n]*h[n-k] $$
Replacing $x[n]$ and $h[n]$
$$y[n] = \sum_{k=-\infty}^\infty cos({\pi}k)*\left(\frac{-1}{2}\right)^{(n-k-2)} $$
Since $cos({\pi}k) = 1$ for k odd and $cos({\pi}k) = -1$ for k even, can I say that $y[n] = 0$ since every odd term on the sum cancels every even term? 
Is that statement above general enough to say that for every LTI system with input signal = $cos({\pi}k)$ the $y[n]$ will always be $0$ no matter $h[n]$?
 A: A well known property of linear time-invariant (LTI) systems is that exponentials are eigenfunctions. That is, if the signal $x[n]=e^{i\omega n},n\in\mathbb{Z}, \omega \in \mathbb{R}$, is input to a stable LTI filter $h[n],n\in\mathbb{Z}$, the output will be equal to 
$$
\begin{align}
y[n]&=\sum_{k\in\mathbb{Z}}h[k]x[n-k]\\
&=e^{i\omega n}H\left(e^{i \omega}\right),
\end{align}
$$
for all $n\in \mathbb{Z}$, where 
$$
\tag{1}
H\left(e^{i \omega}\right)\triangleq \sum_{n\in \mathbb{Z}}h[n]e^{-i\omega n},\omega \in \mathbb{R},
$$
is the, so called, discrete-time Fourier transform (DTFT) of $h[n]$ (assuming that the sum of the RHS of (1) exists).
Now, note that 
$$
\cos(\omega n)=\frac{1}{2}e^{i \omega n} + \frac{1}{2} e^{-i \omega n}.
$$
It is easy to generalize the above and show that the output $y[n]$ with an input $x[n]=\cos(\omega n)$ equals
$$
y[n] = \frac{1}{2}H\left(e^{i \omega}\right) e^{i\omega n} + \frac{1}{2}H\left(e^{-i \omega}\right) e^{-i\omega n}
$$
In general the above output is $not$ equal to zero (for all $n \in \mathbb{Z}$) unless it holds $H\left(e^{i \omega}\right) = H\left(e^{-i \omega}\right)=0$. This conclusion, of course, holds also for the case $\omega = \pi$ as in your question.
Remark: Note that the filter $h[n]$ in your question, does not have a DTFT, hence the above analysis does not hold (unstable filters are not interesting both from a practical and mathematical perspective).
