Find the radius of convergence and interval of convergence of the following power series ∑n=1 ∞ (X-1)^n ÷ 3n(n+1)^3

Question

Find the radius of convergence and interval of convergence of the following power

series

∑_n=1 ^∞ (X-1)^n ÷ 3n(n+1)^3

Please I hope if you know how to answer, when you explain this question to put in mind that I need a little more clarification, than normal students do, thanks in advance.

Answer 1

We using limit ratio test: lim n →∞ of abs(u_{n + 1}/u_n) = lim n →∞ [3n(n+1)³]/[3(n+1)(n + 2)³]·abs(x - 1) = abs(x - 1) < 1 for convergence. So, - 1 < x - 1 < 1 or 0 < x < 2. So radius of convergence is 1.

Now at x = 0 we get u_n = (- 1)ⁿ/3n(n + 1)³ and series converges because abs(u_n) < 1/n⁴(comparison tesr).

The same for x = 2. And because interval of convergence for given series: 0 ≤ x ≤ 2 or [0,2]