This frequency spectrum of a signal $h_2[n]$ is bugging me. I am not sure if what I've done here is correct. It's the sum
\begin{align*} \sum_{n=-\infty}^{\infty}(-1)^{n-1} \end{align*}
in particular I'm struggling with. Afaik this sum is equal to zero but in that case anything of that signal would be zero. Since there are a few other examples which require me to have a correct $H_2(e^{j\theta})$ I'm asking if someone could verify the correctness of this:
\begin{align*} H_2(e^{j\theta}) &= \sum_{n=-\infty}^{\infty} h_2[n] \cdot e^{-jn\theta} \\ &= \sum_{n=-\infty}^{\infty} (0.15 \cdot \delta[n] + 0.3 \cdot \delta[n-1] + 0.15 \cdot \delta[n-2]) (-1)^{n-1} \cdot e^{-jn\theta} \\ &= (0.15 + 0.3 \cdot e^{-j\theta} + 0.15 \cdot e^{-j2\theta}) \cdot \sum_{n=-\infty}^{\infty}(-1)^{n-1} \\ &= 0.3 \cdot (\frac{1}{2} + \cdot e^{-j\theta} + \frac{1}{2} \cdot e^{-j2\theta}) \cdot \sum_{n=-\infty}^{\infty}(-1)^{n-1} \\ &= 0.3 e^{-j\theta} \cdot (1 + \frac{1}{2} ( e^{j\theta} + e^{-j\theta})) \cdot \sum_{n=-\infty}^{\infty}(-1)^{n-1} \\ &= 0.3 e^{-j\theta} \cdot (1 + cos(\theta)) \cdot \sum_{n=-\infty}^{\infty}(-1)^{n-1} \\ \end{align*}
If that is correct, then this should be correct too:
\begin{align*} |H_2(e^{j\theta})| &= \Big|0.3 e^{-j\theta} \cdot (1 + cos(\theta)) \cdot \sum_{n=-\infty}^{\infty}(-1)^{n-1}\Big| \\ &= 0.3 \cdot |(1 + cos(\theta))| \cdot \Big|\sum_{n=-\infty}^{\infty}(-1)^{n-1}\Big| \\ \end{align*}
But since
\begin{align*} \sum_{n=-\infty}^{\infty}(-1)^{n-1} = 0 \end{align*}
what would be the point of this anyway?
I assume that I made a mistake or a false assumption somewhere.
HINT:
Only three elements of the sum are non-zero. So you get a finite sum over just three elements. And by the way, the first sum you're referring does not converge.