I provide my solution to the problem and wonder if I was thinking in a correct way.
Find the radius of convergence of $$\sum_{n=0}{1 \over 1+n3^n}z^n$$ and give with reasoning a point $z_0$ on the edge of the disc at which the series converges.
Here's how I hustle this problem: $$ \begin{eqnarray} \rho=\lim_{n \to \infty}\left|{(1+(n+1)3^{n+1} \over (1+n3^n)} \right| = \lim_{n \to \infty}\left|{1+(n+1)3^{n+1} \over (1+n3^n)} \right|\\ = \lim_{n \to \infty}\left|{1+n3^{n+1}+3^{n+1} \over 1+n3^n} \right|\\ = \lim_{n \to \infty}\left|{{1 \over 3^n}+3n+3 \over {1 \over 3^n}+n} \right|\\ =\lim_{n \to \infty}\left|3+ \frac3n \right|\\ = 3 \end{eqnarray} \\ \text{The series converges, when }|z|<3 \text{ or } -3 < z < 3 $$
Thus we know that the series converges for all $z \in \, ]-3,3 \,[$.
Let's hope one of the edge points $\{3,-3,3i,-3i\}$ will satisfy the convergence condition (in case none of them suits, we need to evaluate something more complicated in form of $x+iy$). Using ratio test we check $$ \begin{eqnarray} \text{for } z_0 = 3:\qquad \rho =\lim_{n \to \infty}{3^n \lbrack 1+(n+1)3^{n+1}\rbrack\over 3^{n+1}(1+n3^n)} = \lim_{n \to \infty}{1+(n+1)3^{n+1}\over3+n3^{n+1}}\\ =\lim_{n \to \infty}{n+1\over n}=1 \end{eqnarray} $$ As $|z_0|=3>\rho=1$, the series diverges at this point. $$ \text{for } z_0 = -3: \qquad \rho =\lim_{n \to \infty}\left|{(-3)^n \lbrack 1+(n+1)3^{n+1}\rbrack\over (-3)^{n+1}(1+n3^n)}\right|= 1 $$ As $|z_0|=3>\rho=1$, the series diverges at this point.
However having made this step I decided to make sure I am doing well and use Mathematica to find out if the series really doesn't converge at $-3$ and $3$. Mathematica told me, that it doesn't converge at $3$, however it does at $-3$, which contradicts my calculations.
SumConvergence[((-3)^ n)/(1 + 3^n*n), n]
True
Where have I mistaken?
For $z=-3$ you can use simply Leibniz criterion: you have convergence of the series since $\frac{3^n}{1+n3^n}$ converges to zero.
For $z=3$ the simplest way is the comparison test: $\frac{3^n}{1+n3^n}\geq\frac{1}{1+n}$. Since the harmonic series diverges also your series does.