Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

tunahan

  • 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
  1. cinar
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. tunahan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that means my solution is nearly correct right?

  3. cinar
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. tunahan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why we get Radius 1/L ?

  5. cinar
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it is definition..

  6. tunahan
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm ok thank you cinar

  7. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy