What's wrong with my radius of convergence test? Given $\sum \limits _{n=2} ^\infty \frac{(\ln x)^n} n$, find its radius of convergence $R$.
Using the ratio test, I arrived at 
$$|\ln x| < 1 \implies e^{|\ln x|} < e^x \implies |x| < e \implies R = e .$$
This isn't the case, however. My teacher did:
$$|\ln x| < 1 \implies -1 < \ln x < 1 e^ \implies {-1} < x < e^1 \implies R = \frac{e-e^{-1}} 2 .$$
I figure that for the first discrepancy (where we decided to take care of the absolute value), I can't do one of the operations I did with an absoulute value sign. My best guess is that I can't cancel $e^{|\ln x|}$, although I'm not sure why -- I am looking for confirmation on if this is indeed the problem and why (or if the problem is somewhere else, where and why).
Secondly, how should I interpret the result? If $\ln x$ is only defined for $x>0$, then theoretically neither radius of convergence (wrong or right one) should work ($-\frac{e-e^{-1}} 2$ is a negative number). So how does this work out?
 A: Where do you see a negative argument for the logarithm? The fact is logarithm is defined only for poitive numbers, but its values can be negative, so:
$$|\log x|<1\iff -1<\log x<1$$
Observe that you have no problem at all applying the exponential now as it is defined in the whole real line. Also, use that $\;e^x\;$ is monotonic ascending:
$$-1<\log x<1\implies e^{-1}<x<e\implies R=\frac{e-e^{-1}}2$$
A: Your error resides in the fact that, unlike what you were expecting, in reality $\textrm e ^{| \ln x |} \ne | \textrm e ^{\ln x} |$. To better understand why, choose $x = \frac 1 {\textrm e}$ and convince yourself that $\textrm e ^1 \ne |\textrm e ^{-1}| = \frac 1 {\textrm e}$.
To answer your second question, notice that your series is not a true power series in $x$. Letting $y = \ln x$, though, turns it into a power series in $y$, with the radius as computed by your professor.
To understand what this tells about $x$, remeber that $\ln$ is defined only for $x>0$, so once you solve the inequation $-R < \ln x < R$ for $x$, don't forget to intersect the solution set with $(0, \infty)$ to get the correct solution for $x$.
To conclude, your suspicions were correct, and you were pretty close to answer your question yourself, which is very good.
