The series $1-1+1-1+\dots$

Question

$1+x+x^2+\dots=\frac{1}{1-x}$, where $-1<x<1$

But I saw in one video this expression is true for $x=-1$ too. So can I write $1+x+x^2+\dots=\frac{1}{1-x}$, where $-1\le x<1$?

Answer 1

No. The geometric seies $1+x+x^2+\dots$ coverges $ \iff -1 <x <1$. In this case we have

$1+x+x^2+\dots=\frac{1}{1-x}$.

For $x=-1$ the series is divergent.

Answer 2

What we mean by $1+x+x^2+\cdots$ is the limit of the partial sums $a_n=\sum_{r=0}^n x^r$. When $-1<x<1$ then it is possible to show that this limit exists and equals $\frac 1{1-x}$. But when $x=-1$ we ahve $a_n=1$ if $n$ is even, and $a_n=0$ if $n$ is odd, so they don't tend to a limit.