Full disclosure: This is a homework problem, but my question is regarding a concept that came about during solving the problem, not the actual solution to the problem.
Problem: Rewrite as geometric power series (might not be correct formulation of problem, I'm still learning :] ) $$\sum_{n=0}^{\infty} (\ln x)^n$$
Formulating this as a geometric series with $a_1=1$ and $r=\ln x$, it converges to $$S=\frac{1}{1-\ln x}$$ if $|\ln x| < 1$.
I rewrite the above inequality to get rid of the absolute value as $-1 < \ln x < 1$.
Now, when I exponentiate (to undo the log), I get $$e^{-1} < x < e^1$$
The question: If the interval contained negative numbers, would I truncate it just to contain positive numbers as the argument of a log cannot be negative?
Do you mean your final interval? That can't happen, since a positive number raised to any power (positive or negative) is positive. (After all, that's why the argument of a log can't be negative.)