Find the radius and interval of convergence of ∑ 0,∞ (n/3^(2n-1))(x-1)^(2n)

Use the ratio test for power series: the ratio

|a

_{n+1}/a_{n}| → q < 1 as n→∞ is sufficient (but not necessary) for convergence.Therefore, we consider

|a

_{n+1}/a_{n}| = |(n+1)/n * (3^{2n+1}/3^{2n-1})*(x-1)^{2n+2}/(x-1)^{2n}|=|(n+1)/n * (3

^{2})*(x-1)^{2}| → |3^{2}(x-1)^{2}| as n→∞For convergence,

|3

^{2}(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/3^{4n-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].