Author Topic: TT1 Problem 2 (night)  (Read 973 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2446
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
TT1 Problem 2 (night)
« on: October 19, 2018, 04:13:20 AM »
Determine the radius of convergence

(a)  $\displaystyle{\sum_{n=1}^\infty \frac{z^n}{2^n n^2}}$

(b) $\displaystyle{\sum_{n=1}^\infty  \frac{z^{3n} (3n)!}{20^n (2n)! }}$

If the radius of convergence is $R$, $0<R< \infty$, determine for each  $z\colon |z|=R$ if this series converges.

Heng Kan

  • Full Member
  • ***
  • Posts: 15
  • Karma: 18
    • View Profile
Re: TT1 Problem 2 (night)
« Reply #1 on: October 19, 2018, 09:34:01 AM »
See the attached scanned picture.

Xiting Kuang

  • Jr. Member
  • **
  • Posts: 7
  • Karma: 8
    • View Profile
Re: TT1 Problem 2 (night)
« Reply #2 on: October 19, 2018, 09:37:28 AM »
Just a concern, it says in the problem that R should be positive.

Heng Kan

  • Full Member
  • ***
  • Posts: 15
  • Karma: 18
    • View Profile
Re: TT1 Problem 2 (night)
« Reply #3 on: October 19, 2018, 09:45:26 AM »
I think the question means that if the radius of convergence is positive,you have to figure out whether the series is convergent at the radius of convergence. It doesn't mean the radius is always positive.

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2446
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: TT1 Problem 2 (night)
« Reply #4 on: October 20, 2018, 03:20:20 PM »
I think the question means that if the radius of convergence is positive,you have to figure out whether the series is convergent at the radius of convergence. It doesn't mean the radius is always positive.
Indeed.