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

Question

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

series

∞

∑ (X-1)^n ÷( 3n(n+1)^3)

n=1

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

Taha:

Let a_{n =} (x-1)ⁿ/[3*n*(n+1)³]

a_n+1 = (x-1)⁽ⁿ⁺¹⁾/[3*(n+1)*(n+2)³]

We need to determine | a_n+1 / a_n |

| a_n+1 / a_n | = | [(x-1)ⁿ⁺¹*3*n*(n+1)³] / [3*(n+1)*(n+2)³*(x-1)ⁿ] |

| a_n+1 / a_n | = | (x-1) * (n/n+1)*((n+1)/(n+2))³ |

as n goes to ∞ | a_n+1 / a_n | approaches |x-1|

To determine the interval of convergence, determine where |x-1|<1 so -1<x-1<1

Taha, don't believe you are bad at math! Power series are tricky

So we know the interval of convergence is at least 0<x<2

We need to test the endpoints x=0 and x=2

When x = 0 We have ∑ (-1)^n / (3*n*(n+1)^3)

By alternating series test, we know this converges since the limit as n goes to ∞ of 1/(3*n*(n+1)^3) = 0 and that 1/(3*n*(n+1)^3) is a decreasing function.

For x = 1, we have ∑ 1 / (3*n*(n+1)^3)

1 / (3*n*(n+1)^3) = 1/(3n^4) when n is large, so ∑ 1 / (3*n*(n+1)^3) behaves like ∑ 1/(3n^4) when n is large. 1/(3n^4) is a p-series with p = 4 > 1 so it converges. So ∑ 1 / (3*n*(n+1)^3) also converges.

So our interval of convergence is x is in [0,2] we include the endpoints.