Power series: Is the radius of convergence $\frac{1}{3}$ for $ \frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3}+...$ I should calculate the radius of convergenc and would like to know, if the result $\frac{1}{3}$ is correct.
Here the exercise:

$$ \frac{x}{1\cdot3} + \frac{x^2}{2\cdot3^2} + \frac{x^3}{3\cdot3^3} + \frac{x^4}{4\cdot3^4}... $$

This is:
$$
\sum\limits_{n=0}^\infty \frac{x^{n+1}}{(n+1)3^{n+1}} \\
\lim\limits_{n \to \infty} \left| \frac{(n+1)\cdot3^{n+1}}{(n+2)\cdot3^{n+2}} \right| = \left| \frac{1}{3} \right|
$$
I’m right? Thanks.
Summery
I could test with the ratio test if a power series is convergent. 
I could use $$\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|}$$ and get the $\left|x\right|$ for which the series is convergent. With that test the series is convergent, if the result is $<1$.
 A: Using the ratio test for absolute convergence.
$$
|a_{n+1}| = \frac{|x|^{n+2}}{(n+2)3^{n+2}} \\
$$
$$
|a_{n}| = \frac{|x|^{n+1}}{(n+1)3^{n+1}} \\
$$
$$
\frac{|a_{n+1}|}{\left|a_{n}\right|} = |x| \left( \frac{n+1}{n+2} \right)\left( \frac{1}{3} \right)
$$
$$
\lim\limits_{n \to \infty} \frac{|a_{n+1}|}{\left|a_{n}\right|} = \frac{|x|}{3}
$$
The series converges absolutely if $\frac{|x|}{3} < 1 $, which is when $|x| < 3$. Absolute convergence implies convergence.
You also need to check for convergence when $|x| = 3$ to determine if those points are in the radius of convergence.
A: The easiest way to check the convergence of $a_0 + a_1(z-\omega) + a_2(z-\omega)^2 + \cdots$ is to apply the ratio test. To apply the ratio test, we need
$\lim_{k \to \infty} \frac{|a_{k+1}(z-\omega)^{k+1}|}{|a_k(z-\omega)^k|} < 1 \, .$
Simple algebraic manipulations show us that
$\lim_{k \to \infty} \frac{|a_{k+1}(z-\omega)^{k+1}|}{|a_k(z-\omega)^k|} < 1 \iff |z-\omega| < \lim_{k \to \infty}\left| \frac{a_k}{a_{k+1}} \right| . $
In other words, if we denote the far right hand limit (assuming it exists) by $\rho$, then the ratio test tells us that the series converges for all $z$ within a distance of $\rho$ of $\omega,$ i.e. $|z-\omega| < \rho.$
I think your problem was that you had the $a_k$ and $a_{k+1}$ the wrong way up and you got the reciprocal of the radius of convergence. In your case $\omega = 0$ and $a_k = 1/(k+1)3^{k+1}$, as you have correctly identified. It follows that
$\rho = \lim_{k\to\infty} \left| \frac{(k+2)3^{k+2}}{(k+1)3^{k+1}} \right| = 3\lim_{k \to \infty} \left|\frac{k+2}{k+1} \right| = 3 \, .$
Thus, your series converges for all $|z| < 3.$
