Search 73,256 tutors
FIND TUTORS
Ask a question
0 0

Find the radius and interval of convergence

Tutors, please sign in to answer this question.

1 Answer

Use the ratio test for power series: the ratio
 |an+1/an| → q < 1 as n→∞ is sufficient (but not necessary) for convergence.
Therefore, we consider
|an+1/an| = |(n+1)/n * (32n+1/32n-1)*(x-1)2n+2/(x-1)2n|
=|(n+1)/n * (32)*(x-1)2|  → |32 (x-1)2| as n→∞
For convergence,
|32 (x-1)2| < 1
|x-1| < 1/3
 
Therefore, the radius of convergence is 1/3.
The interval of convergence is at least |x-1| < 1/3, or 2/3 < x <4/3, but we need to check convergence at the interval endpoints: Since (x-1)2n=(1/3)2n at both endpoints, the series is ∑n=1 n/34n-1, which converges by the ratio test. Therefore, the interval of convergence is |x-1| ≤ 1/3, or 2/3 ≤ x ≤ 4/3, or [2/3, 4/3].