Question

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?

Answer 1

\[|R|=3 \quad -3\le R \le 3\]

Answer 2

that means my solution is nearly correct right?

Answer 3

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

Answer 4

why we get Radius 1/L ?

Answer 5

it is definition..

Answer 6

hmm ok thank you cinar