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

Question

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

series

∑_n=1^∞ (X-1)^n ÷ (3^n) *(n+1)^3

PLEASE SOLVE WITH STEPS

Answer 1

Taha:

1) let a_n = (x-1)ⁿ/(3ⁿ*(n+1)³)

2) determine a_n+1 and then a_n+1/a_n

a_n+1 = (x-1)ⁿ⁺¹/(3ⁿ⁺¹ *(n+2)³)

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

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

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

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

3) determine | a_n+1/a_n | and lim n→∞ |a_n+1/a_n|

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

lim n→∞ |a_n+1/a_n| = |(x-1)/3| since lim n→∞ ((n+1)/(n+2))³ = 1

4) Determine where lim n→∞ |a_n+1/a_n| < 1

|(x-1)/3| <1 implies |x-1|<3 Radius of convergence is 3

or -3 < x-1< 4

or -2 < x < 4 We know we have convergence on at least -2 < x < 4

5) Test the endpoints x= -2 and x=4

Plug x= -2 into ∑(x-1)ⁿ/(3ⁿ*(n+1)³) to get ∑ (-1)ⁿ/ (n+1)³

Use alternating series test to show ∑ (-1)ⁿ/ (n+1)³ converges or diverges. To show convergence, show 1/(n+1)³ is decreasing and show lim n →∞ 1/(n+1).

I'll let you show this.

Plug x = 4 into ∑(x-1)ⁿ/(3ⁿ*(n+1)³) to get ∑1/(n+1)³

Use the fact that 1/(n+1)³ ≅ 1/n³ when n is large.

Does ∑1/n³ converge or diverge?

I'll let you figure this out too.

Let me know if you need further help!