Range of values of x for the sum to be valid.

Question

Given the series: $$\frac{1}{x+1}+\frac{1}{(x+1)^2}+\frac{1}{(x+1)^3}+...$$

Find the sum to infinity for the series and state the range(s) of values of $x$ for the sum to be valid.

I solved the first part of question but I've no idea for the 'range' part. Hope someone can point it out. Thanks in advance.

Answer 1

It's a geometric series

For $-1<\frac{1}{x+1}<1$ it converges to $\frac{\frac{1}{x+1}}{1-\frac{1}{x+1}}=\frac{1}{x}.$

The range of $x$ is $$(-\infty,-2)\cup(0,+\infty)$$ because $\frac{1}{x+1}<1$ it's $$\frac{1}{x+1}-1<0$$ or $$\frac{x}{x+1}>0,$$ which gives $x>0$ or $x<-1$;

$$\frac{1}{x+1}>-1$$ it's $$\frac{1}{x+1}+1>0$$ or $$\frac{x+2}{x+1}>0,$$ which gives $x<-2$ or $x>-1$ and which the previous result we get:

$x<-2$ or $x>0$.