## anonymous 3 years ago calculate the radius of convergence and the interval of convergence (with examination of the endpoints) $\sum_{n=1}^{\infty}\frac{3^{n+1}}{n}.x^{n}$ a try... $|\frac{a_{n+1}}{a_{n}}|=$ $\Large \frac{\frac{3^{(n+1)+1}}{n+1}}{\frac{3^{n+1}}{n}}=$ $\frac{3^{n+2}}{n+1}.\frac{n}{3^{n+1}}=\frac{3}{1}=3$ is it correct and what i need to do additional?

1. anonymous

$|R|=3 \quad -3\le R \le 3$

2. anonymous

that means my solution is nearly correct right?

3. anonymous

sorry I made a mistake, first find the Limit, in your case L=3 then Radius is |R|=1/L

4. anonymous

why we get Radius 1/L ?

5. anonymous

it is definition..

6. anonymous

hmm ok thank you cinar