Find the radius of convergence R for the series n = 1 ∑ ∞ ( 2 3 ) n ( n + 1 ) 2 n x n .
Submit your answer as ⌊ 1 0 0 0 R ⌋ .
Bonus : What happens to the series when ∣ x ∣ = R and when x ∈ R ?
Notation : ⌊ ⋅ ⌋ denotes the floor function .
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Yes,that is... and very well explained ↑ , thank you very much, compañero!. I just only wanted to point that we can also use root criterion or Cauchy's criterion ( n → ∞ lim n ∣ a n ∣ ...) for calculating the radius of convergence R of this series, and at x = 3 − 2 we can also use Dirichlet convergence criterion for seeing convergence at this point.
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The radius of convergence of ∑ a n x n is R = lim n → ∞ ∣ ∣ ∣ a n + 1 a n ∣ ∣ ∣ if that limit exists. In our case this is the limit of ( 2 3 ) n ( n + 1 ) 2 n ( 3 2 ) n + 1 n + 1 ( n + 2 ) 2 , which is 3 2 . The answer is 6 6 6 ... always a great answer, compañero ;)
At x = 3 2 the series diverges by comparison with the harmonic series. At x = − 3 2 it converges to ln ( 2 ) − 1 2 π 2 ; write ( n + 1 ) 2 n = n + 1 1 − ( n + 1 ) 2 1 .
I was afraid you would ask about x = 3 2 i ;)