I have a series summation of the form
$$ \tag{1} S(x) = \sum_{i = -\infty}^{\infty} (-1)^{i}\left[\Phi\left(2ix\right) - \Phi\left((2i-2)x\right)\right], $$ where $\Phi(.)$ is the standard normal cumulative distribution function. We can rewrite (1) as
$$ \tag{2} S(x) = 2\Phi(0) + \sum_{i = 1}^{\infty} (-1)^{i}\left[\Phi\left(2ix\right) + \Phi\left(-2ix\right)\right], $$ which transforms into
$$ S(x) = 1 + \left(-2+2-2+2-2\dots\right), $$ which obviously is bounded but divergent.
But another way to write $S(x)$ is
$$ \tag{3} S(x) = \lim_{M \rightarrow \infty} \sum_{i \in [-M,M]} (-1)^{i}\left[\Phi\left(2ix\right) - \Phi\left((2i-2)x\right)\right]. $$
I know that it is mathematically incorrect (even blasphemous) to suggest that $-\infty = (-1)\times \infty$, but since the summation limits are from $-M$ to $M$, it intuitively seems correct to assume that the summation limits are symmetric over the origin. As long as the summation limits are symmetric, $S(x)$ becomes
$$ \tag{4} S(x) = 2\Phi(0)- \lim_{M\rightarrow -\infty} \Phi((2M-2)x), $$ which is 1.
I was suggested to explore Cesaro Summation. Since all iterations are equal to 0 in the way the summation is written, it seems that I can claim $S(x)=1$.
However, I know that life is not so easy. So, my main question is what kind of logical (or mathematical) fallacy I commit here? Also, where is the logical fallacy? Is it in the transition from (1) to (3) or in the evaluation process in (4)?
My final question is, from an applied mathematics perspective, can I assume that $S(x)=1$? If I can, what is the justification?
Thanks in advance!
There is an unnecessary step "We can rewrite (1) as..." which only complicate things. Series (1) converges absolutely for any $x\in \mathbb R$ and, therefore, can be summed up to the same value $S(x)$ rearranging the terms in any order. So indeed (3) holds.