Series' convergence - making my ideas formal 
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges. 

My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \infty} \dfrac{3^{n+1}|x|+n|x|+|x| }{3^n+n} = \lim_{n \to \infty} ( \dfrac{3^{n+1}|x|}{3^n+n} + \dfrac{n|x|}{3^n+n} + \dfrac{|x|}{3^n+n}) $$
$$ = \lim_{n \to \infty} (\dfrac{3|x|}{1+\dfrac{n}{3^n}} + \dfrac{|x|}{\dfrac{3^n}{n} + 1} + \dfrac{|x|}{3^n + n}) = 3|x| + 0 + 0 = 3|x| $$
So we want $3|x| < 1$, i.e. $|x| < \dfrac{1}{3}$. But we have to also check the points $x=\dfrac{1}{3}, -\dfrac{1}{3}$, where the limit equals $1$. 
For $x=\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty (1 + n \cdot \dfrac{1}{3}^n)  $ which obviously diverges.
For $x=-\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (-\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty ((-1)^{n} + n \cdot (-\dfrac{1}{3})^n)$. Now I know this diverges because $(-1)^n$ and $(-\dfrac{1}{3})^n$ have alternating coefficients. But is there a theorem that I can use here? 
So we conclude that $|x| < \dfrac{1}{3}$, but is the above work enough to show it conclusively? 

The series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges if $|x|<\frac{1}{3}$
Indeed $$(3^n + n)\cdot x^n\sim_{\infty} 3^n \cdot x^n$$ but this is a geometric series that converges only if $|x|<\frac {1}{3}$
