The sequence converges, since each term is #-1/4# times the previous term, and #abs(-1/4) < 1#. Thus the infinite sum exists. Let us call this sum #S#. Then:

#color(white)(–1/2)S=-2+1/2-1/8+...#

#color(white)(–1/2S)=-2(1-1/4+1/16-1/64+...)#

#color(blue)(–1/2 S=" "1-1/4+1/16-1/64+...)#

If we multiply both sides by #-1/4#, we get

#(–1/4)(–1/2 S)=(–1/4)(1-1/4+1/16-1/64+...)#

#color(green)(" "1/8 S = -1/4 + 1/16 - 1/64 + ...)#

Notice how the expansion for #color(blue)(-1/2 S)# includes the expansion for #color(green)(1/8 S)#. If we subtract #1/8 S# from #-1/2 S# we get:

#" "color(blue)(–1/2 S) - color(green)(1/8 S)=color(blue)(" "1-1/4+1/16-1/64+...)#

#color(white)(–1/2 S - 1/8 S=)color(green)(" "-(-1/4+1/16-1/64+...))#

#" "–5/8 S=1#

Solving for #S# gives:

#S=-8/5#