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$$y[n]=(-1)^n w[n]+w[n]$$
$n$ ranges from $-\infty$ to $\infty$
where $w[n]$ is output of frequency response block with cutoff frequency = $\pi/2$, and input is unit impulse function.
I know output of frequency response block is $w[n]$ is $h[n]$ because $\delta[n]$ conv with $h[n]$ = $h[n]$ since delta exists only at zero
May be we can solve this in frequency domain?
The half-band lowpass filter, has the following impulse response : $$ h[n] = \frac{\sin(\frac{\pi}{2} n) }{ \pi n} $$
And for $x[n] = \delta[n]$ , $w[n] = h[n]$.
Then the output $y[n]$ will be :
$$y[n] = w[n] \big( 1 + (-1)^n \big) $$
which can be simplified by expanding the parenthesis as:
$$ \begin{align} y[n] &= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot \big( e^{j 2 \pi n} + e^{j \pi n} \big) \\ \\ y[n] &= \frac{\sin(\frac{\pi}{2} n) }{ \pi n} \cdot e^{j \frac{3 \pi}{2} n} \cdot \big( e^{j \frac{\pi}{2} n} + e^{-j \frac{\pi}{2} n} \big)\\ \\ y[n] &= \frac{ \sin(\frac{\pi}{2} n) }{ \pi n} \cdot 2 \cdot \cos( \frac{\pi}{2} n) \cdot e^{j \frac{3 \pi}{2} n} \\ \\ y[n] &= \frac{\sin(\pi n) }{ \pi n} \cdot e^{j \frac{3 \pi}{2} n} \end{align} $$ Which can be shown to be :
$$y[n] = \begin{cases}{ 1 ~~~~, n = 0 \\ 0 ~~~~ , n \neq 0 } \end{cases} $$
Therefore $y[n] = \delta[n]$. The system is a zero-phase all pass filter. And finally, the required quantity is $$ M = \sum_{n=-\infty}^{\infty} y[n] = \sum_{n=-\infty}^{\infty} \delta[n] = 1 $$
