$\lim_{n \to \infty} \frac{(1-2+3-4+5-...-2n)}{(1+n^2)^.5}$
can be solved as $\lim_{n\to\infty}\frac{-n}{(1+n^2)^.5}$
$\lim_{n\to\infty} \frac{-n}{n(1/n^2+1)^.5}$
$=-1$
my question is why can't the numerator be equal to $1/4$ as it forms Grandi Series?
The technical calculations you wrote are correct and arrive at the the correct result.
Your question about the Grandi series seems to imply that what you want to do is use the rule about the limit of a quotient being the quotient of the limits:
$$\lim_{n \to \infty} \frac{(1-2+3-4\pm\ldots-2n)}{(1+n^2)^.5} = \frac{\lim_{n \to \infty} (1-2+3-4\pm\ldots-2n)}{ \lim_{n \to \infty}(1+n^2)^.5}.$$
That rule however works only if both limits on the right hand side exist and the limit of the denominator is not $0$.
In this case we have the enumerator (just $-n$) diverge to $-\infty$ and the denominator diverge to $+\infty$, so this rule can't be applied.
Even if you consider the enumerator as the series
$$\sum_{n=1}^{\infty}(-1)^{n+1}n = 1-2+3-4\pm \ldots$$
the series of partal sums would alternate between $-\frac k2$ if taken to an $k$ even and $\frac{k+1}2$ if taken to an odd $k$, which is also divergent (thanks at J.G. to pointing that out).