I know that this series is not summable in the usual sense so I'm "playing the game" of divergent series summation. The series is:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})=1-2+1-2+1-2+\dots$$
I tried Abel summation method which failed this cause:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=1-2x+x^2-2x^3+\dots=\frac {1-2x}{1-x^2}$$
Has singularities at $x=\pm 1$
Then i tried Borel summation but again I get to:
$$\sum_{n=0}^{+\infty}-\frac 12(1+3(-1)^{n+1})x^n=\frac 12(\frac 1{x-1}-\frac 3{x+1})$$
For $x\in ]-1;+\infty[-\{1\}$ that gives the same problem, and I don't think this is the case for Ramanujan summation.
I started to try Cesaro summation and I got to the following sequence of partial sums:
$$1;-1;0;-2;-1;-3;-2;-4;\dots$$
And the sequence of Cesaro means seems to diverge:
$$1;0;0;-\frac 12;-\frac 35;-1;-\frac 87;-\frac 32;\dots$$
I really don't know what other method should I try.
You might try Zeta function regularization.
EDIT: Warning: depending on how you do this, you can get different results. One way is this:
$$\sum_{n=1}^\infty -\frac{1}{2} (1 + 3 (-1)^n) n^{-s} =\zeta(s) - \dfrac{3}{2^s} \zeta(s)$$ and taking $s=0$ gives us the value $1$.