Taha T.

asked • 05/30/20# Find 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 a_{n} = (x-1)^{n}/(3^{n}*(n+1)^{3})

2) determine a_{n+1} and then a_{n+1}/a_{n}

a_{n+1 }= (x-1)^{n+1}/(3^{n+1} *(n+2)^{3})

a_{n+1}/a_{n }= [(x-1)^{n+1} * 3^{n} * (n+1)^{3}] / [(x-1)^{n} * 3^{n+1}*(n+2)^{3}]

a_{n+1}/a_{n }= [(x-1)^{n+1}/(x-1)^{n}] * [3^{n }/ 3^{n+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}/(3^{n}*(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}/(3^{n}*(n+1)^{3}) to get ∑1/(n+1)^{3}

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

Does ∑1/n^{3} 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