$$s_n=\sum_{k=0}^n \left(\frac{-1}{4}\right)^k$$
I thought about interpreting this as a geometric progression, however, i am really unsure about using it with an negative base. If this okay, this sum would converge as following:
$$s_n=\frac{1-(-0,25)^{n+1}}{1-(-0.25)}\xrightarrow[n\to\infty]{}\frac{1}{1.25}$$
Please correct me if I am wrong at this point.
On
If you want to convince yourself that one can use a negative value of $x$ with the same result, you could group the elements two by two in this way $$ (1+x)+(x^2+x^3)+\cdots+(x^{2N}+x^{2N+1})=(1+x)+(1+x)x^2+\cdots+(1+x)x^{2N}$$ or in a more condensed way $$\sum_{n=0}^{2N+1}x^{n}=\sum_{n'=0}^N(1+x)x^{2n'}=(1+x)\sum_{n'=0}^N(x^2)^{n'}.$$ Now notice that the last sum is what you already know with now a positive base $x^2$. So you know that it is equal to $(1-(x^2)^{N+1})/(1-x^2)$. In you expression, you can see then that $$\sum_{n=0}^{2N+1}x^{n}=(1+x)\frac{1-x^{2N+2}}{1-x^2}=\frac{1-x^{2N+2}}{1-x}.$$ If the number of elements in the sum is odd, add the last to the above result $$\sum_{n=0}^{2N+2}x^{n}=\sum_{n=0}^{2N+1}x^{n}+x^{2N+2}=\frac{1-x^{2N+2}}{1-x}+x^{2N+2}=\frac{1-x^{2N+2}+x^{2N+2}-x^{2N+3}}{1-x}=\frac{1-x^{2N+3}}{1-x}.$$ I hope this convinces you that the sign of $x$ does not play any role in the formula.
Don't worry: the equality you gave concerning the parial sum would hold for any value of $x$, not only the $\frac{1}{4}$ you wrote it with. It works in any field (ratinals, reals, compelx numbers, finite fields ...), but in any ring, as long as $1-x$ is invertible, and provided you interpret division by $1-x$ as multiplication by its inverse. In general, you can say $\sum x^k$ will converge to $\frac{1}{1-x}$ if $|x|<1$, diverges to plus infinity for $x\geq1$, and oscillates for $x\leq-1$, diverging in modulus if $x<-1$. The convergence statement holds for generic complex numbers, the diivergence one is specific to reals, and I do not really know what happens with complex numbers with absolute value one or over.