
Christopher J. answered 05/27/20
Berkeley Grad Math Tutor (algebra to calculus)
Taha:
Let an = (x-1)n/[3*n*(n+1)3]
an+1 = (x-1)(n+1)/[3*(n+1)*(n+2)3]
We need to determine | an+1 / an |
| an+1 / an | = | [(x-1)n+1*3*n*(n+1)3] / [3*(n+1)*(n+2)3*(x-1)n] |
| an+1 / an | = | (x-1) * (n/n+1)*((n+1)/(n+2))3 |
as n goes to ∞ | an+1 / an | 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