I read this as ∑{n=[1,∞) }-4(-½)ⁿ⁻¹, in words:
the sum between n=1 and infinity of negative 4 times negative one half to the power of n-1.
If we use S to represent the sum, then S=-(4-2+1-½+¼-...)=-(2+1-½+¼-⅛+...).
Let G=1-½+¼-⅛+... =1/(1-(-½)) a convergent geometric progression.
G=1/(3/2)=⅔.
S=-2-G=-2⅔=-8/3.