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? 

For $x=-1/3$, we can separate the even $n$ and odd $n$ terms and obtaining:
$$A=\sum_{n = 1}^{\infty} \left((-1)^n + n \left(-\frac 1 3\right)^n\right)
=\sum_{k = 1}^{\infty} \left((-1)^{2k} + (2k) \left(-\frac 1 3\right)^{2k}\right)
+\sum_{k = 1}^{\infty} \left((-1)^{2k+1} + (2k+1) \left(-\frac 1 3\right)^{2k+1}\right)$$
$$A=\sum_{k = 1}^{\infty} \left(1+ (2k) \left(\frac 1 3\right)^{2k}-1 - (2k+1) \left(\frac 1 3\right)^{2k+1}\right)=\sum_{k = 1}^{\infty} \left((2k) \left(\frac 1 3\right)^{2k}- (2k+1) \left(\frac 1 3\right)^{2k+1}\right)=\sum_{k = 1}^{\infty} \left(\frac 1 3\right)^{2k+1}\left(4k-1\right)$$
You can now apply the ratio test to the last series.
