
Taha T.
asked 05/30/20Find the radius of convergence and interval of convergence of the following power series ∑n=1 ∞ (X-1)^n ÷ (3^n) *(n+1)^3
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
1 Expert Answer

Christopher J. answered 05/31/20
Berkeley Grad Math Tutor (algebra to calculus)
Taha:
1) let an = (x-1)n/(3n*(n+1)3)
2) determine an+1 and then an+1/an
an+1 = (x-1)n+1/(3n+1 *(n+2)3)
an+1/an = [(x-1)n+1 * 3n * (n+1)3] / [(x-1)n * 3n+1*(n+2)3]
an+1/an = [(x-1)n+1/(x-1)n] * [3n / 3n+1 ] * [ (n+1)3/(n+2)3]
an+1/an = [(x-1)n+1-n]*[3n-n-1]*[(n+1)/(n+2)]3
an+1/an = [(x-1)] * [1/3]* [(n+1)/(n+2)]3
3) determine | an+1/an | and lim n→∞ |an+1/an|
| an+1/an | = | (1/3)*((n+1)/(n+2))3* (x-1) |
lim n→∞ |an+1/an| = |(x-1)/3| since lim n→∞ ((n+1)/(n+2))3 = 1
4) Determine where lim n→∞ |an+1/an| < 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!
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Paul M.
05/30/20