Christopher J. answered • 05/27/20

Berkeley Grad Math Tutor (algebra to calculus)

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.

Taha T.

could you please write the steps to the rest of the solution??05/27/20