For what values of x does the series $\sum_{n=1}^{\infty} \frac{n^x}{x^n} $ converge? 
For what values of x does the series $\sum_{n=1}^{\infty}\frac{n^x}{x^n} $ converge?

I've attempted to solve this problem but I can't finish up my reasoning - I don't know how to "check" the remaining numbers. Namely: 
(1) I showed that this series converges absolutely for $x \in (-\infty,-1) \cup (1, \infty)$ 
(2) Then, I checked $x=1$ and $x=-1$ - the series does not converge, because $\frac{n^x}{x^n} $ does not approach $0$. 
(3) Zero does not work because the series is not defined. 
(4). Now, what I am left with is to check the interval $(-1,1)$, but I don't know how to do this. I can cater for $(0,1)$ using the ratio test for the initial series, but what about $(-1,0)$?

For $x \in (-1, 1), x\neq 0$ we have $x^n \overset{n\to\infty}{\longrightarrow} 0$. So for $x>0$ , we know $\frac{n^x}{x^n}$ does not converge to zero.
For $x<0$ we can write $x=-\frac{1}{y}$ with $y>1$. Then the absolute value of our sequence is
$$\lvert\frac{n^x}{x^n}\rvert = \frac{y^n}{n^\frac1y} \ge \frac{y^n}{n}.$$
To see that this doesn't converge to zero, we can take the $\log$:
$$\log \left( \frac{y^n}{n} \right) = n\log(y) - \log(n)> 0 $$ for $n$ big enough. 
That means $\frac{y^n}{n}>1$.
Alternative: (for showing $\frac{y^n}{n}$ doesn't converge to zero)
Recall that $y>1$, so we can write $y=1+ \epsilon$ with $\epsilon > 0$. Now
$$\frac{y^n}{n}=\frac{(1+\epsilon)^n}{n}=\frac{\sum_{k=0}^{n}\binom{n}{k}\epsilon^k}{n}=\frac{1+n\epsilon+\sum_{k=2}^{n}\binom{n}{k}\epsilon^k}{n} > \epsilon$$
