A community for students.
Here's the question you clicked on:
 0 viewing
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?
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?

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[R=3 \quad 3\le R \le 3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that means my solution is nearly correct right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry I made a mistake, first find the Limit, in your case L=3 then Radius is R=1/L

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why we get Radius 1/L ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm ok thank you cinar
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.