<s>[INST] 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 $<1$.
 [/INST] 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.
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