How can I prove the values of convergence for $x^n/n$ without using the Ratio or Root test? If I'm given $\sum \frac{x^n}{n}$ how can I find which values of x the series converges for? 
My gut instinct is that it would be for any values in $-1< x < 1$ because the numerator is an exponential function and the denominator is just n, thus having $|x| < 1$ means the numerator wouldn't go to infinity, but would continue getting smaller as n approaches infinity, thus making the series converge to $0$. 
However, I can't for the life of me figure out how to prove that without using the Ratio test, which is required for the problem. Is this possible and what should I be looking at to prove it without that? It's supposed to be have two different sets of values for x to where it's converging absolutely and converging conditionally.
 A: Your gut instinct is good.  You can just use the comparison test to a geometric sequence,  taking absolute values to show absolute convergence.
If $x \in (-1,1)$, then $|x^n/n|<|x^n|$,  and that's a geometric series which converges, therefore your series converges.  At +1, you have the harmonic series which diverges.  At -1 you have the alternating harmonic series, which converges by the alternating series test.
If $x>1$ or $x<-1$,  then the sequence does not go to 0, so it diverges by the divergence test
A: We have the following cases:


*

*If $x=1$, then $\sum x_n/n=\sum 1/n$ is the harmonic series, which is known to diverge.

*If $|x|>1$, then $x^n/n$ diverges (diverging exponential growth divided by linear growth). For a series $\sum a_n$ to converge it is a necessary criterion that $a_n\to 0$. Hence the series $\sum x_n/n$ cannot converge.

*If $x\in[0,1)$, then the sum is monotone and also dominated by $\sum x^n$ which converges to $x/(1-x)$ for all $x\in(-1,1)$ (geometric series). Hence, $\sum x_n/n$ also converges.

*If $x\in[-1,0)$, then $x^n/n$ is an alternating sequence which converges monotonically to zero. It is known that series over alternating and vanishing sequences always converge.
So over all, your instinct was right and the sequence converges for $|x|<1$ and diverges for $|x|>1$. Additionally, the sequence converges in one of the boundary points of its convergence interval, namely $x=-1$, and diverges in the other one.

By the way, here is some taylor series magic. Lets differentiate the series term-wise:
$$\sum_{n=1}^\infty\frac{x^n}n\quad\to\quad\sum_{n=1}^\infty \frac{nx^{n-1}}{n}=\sum_{n=1}^\infty x^{n-1}=\sum_{n=0}^\infty x^n=\frac1{1-x}.$$
So the derivative seems to be $1/(1-x)$, which implies that $\sum x^n/n$ converges to $-\log(1-x)$ (wherever it converges). However, a rigirous reasoning needs some more thought on taylor series and differentiability.
