How to evaluate the inequality $|x+1|<-1$?

Okay perhaps the title isn't specific enough, I didn't know how to word it exactly. I'm finding the interval of convergence for a power series and i know the answer to be (-2,0]

I end up with the inequality $|x+1|<-1$, my question is how do I resolve the interval (-2,0] from that, I'm sure the answer is probably easier than I'm making it out to be?

Would I make it: $-1<x+1<1$ and than subtract 1 from both sides? If yes, how come I could put negative 1 on the side which makes it less than $x+1$, when previously it was greater?

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I suspect you've made a mistake elsewhere in your analysis. There is, in fact, no value of $x$ for which $|x+1|<-1$, since the absolute value function is non-negative. –  Michael Grant Apr 8 '13 at 5:08
-1 will never be greater than $|.|$ –  SyntacticSugar Apr 8 '13 at 5:09
thanks guys this helped me solve the problem!:) –  lawlipop Apr 8 '13 at 5:27

If $x \in \mathbb{C}$(including $x \in \mathbb{R}$ ofcourse) the $|x+1|$ number is always non-negative(means $\ge 0$).
So you cannot find non-negative number equals to negative($\lt 0$) number.
$|x|$ , where $x$ belongs to $R$ can never be negative.
Hence the answer to the question is $\phi$ or 'Null Set'.