Period of Σ(−1)^k δ[n−k]

$$x[n] = \sum_{k=-\infty}^{\infty}(-1)^k\delta[n-k]$$

Each impulse $\delta[n-k]$ is non-zero only at $n = k$, so at each $n$ exactly one term survives, with value $(-1)^n$:

$$x[n] = (-1)^n = \cos(\pi n) = \{\ldots, 1, -1, \underline{1}, -1, 1, \ldots\}$$

$x[n+2] = (-1)^{n+2} = (-1)^n = x[n]$, and $x[n+1] = -x[n] \neq x[n]$.

(Also $\omega_0 = \pi$, $N = 2\pi m/\pi = 2m$, smallest $N = 2$.)

Answer: fundamental period $N = 2$.