One minus one plus one minus one

You can't complete this summation. A geometric series is only defined if $|r|<1,$ and not if $|r|\ge1.$ This formula can be manipulated to equal any number: $1,$ $42,$ $\pi$ and every other real number.

The summation is $\boxed{\text{undefined}}$

answer will be either 1 or zero depending upon the numbers of terms are even or odd. g.p. for infinite terms is valid only for |r|<1.

It's Grandi's series. it's divergent series because it's partial sums are not approaching a fixed number (they are 0 and 1) but it has cesaro or {R}summation as following

let x = 1-1+1-1+1..... {R}

x= 1-(1-1+1-1......)

x=1-x

2x=1

x=1/2

{R} indicates Ramanujan's summation

Let S=1-1+1-1+1-1+1-1.......

1-S=1-(1-1+1-1+1-1+1-1.......)

1-S=1-1+1-1+1-1+1-1.......

1-S=S

therefore, 2S=1

$$ The infinite series of $$ $\sum_{n=0}^\infty (-1)^n=1-1+1-1+\cdots$ $$ is called Grandi's series . Actually, one can arrive at two conclusions: Grandi's series is undefined or Grandi's series is equal to $\dfrac{1}{2}$ . Both of the conclusions can be correct and formally proven. $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$

sum of a Geometric Progression upto infinity

See this Video... https://www.youtube.com/watch?v=PCu_BNNI5x4