Power Series Convergence Test - Radius of 2/3

[513501265]

Homework Statement

The coefficients of the power series the sum from n=0 to infinity of an (x-2)^n satisfy ao=5 and an= [(2n+1)/(3n-1)] an-1 for all n is greater than or equal to 1. The radius of convergence of the series is

A) 0 B) 2/3 C) 3/2 D) 2 E) infinite

Homework Equations

The Attempt at a Solution

Convergence test: lim n→∞|an (x-2)^n / an+1 (x-2)^n+1|
2/3*|1/(x-2)|<1
x belongs to (4/3, 8/3)
Therefore, the radius is 2/3

 

[LCKurtz]

Homework Statement

The coefficients of the power series the sum from n=0 to infinity of an (x-2)^n satisfy ao=5 and an= [(2n+1)/(3n-1)] an-1 for all n is greater than or equal to 1. The radius of convergence of the series is

A) 0 B) 2/3 C) 3/2 D) 2 E) infinite

Homework Equations

Convergence test: lim n→∞|an (x-2)^n / an+1 (x-2)^n+1|
2/3*|1/(x-2)|<1
x belongs to (4/3, 8/3)
Therefore, the radius is 2/3

For the ratio test you use the n+1 term over the n term ( or n over n-1). You have it upside down.

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

The coefficients of the power series the sum from n=0 to infinity of an (x-2)^n satisfy ao=5 and an= [(2n+1)/(3n-1)] an-1 for all n is greater than or equal to 1. The radius of convergence of the series is

A) 0 B) 2/3 C) 3/2 D) 2 E) infinite

Homework Equations

The Attempt at a Solution

Convergence test: lim n→∞|an (x-2)^n / an+1 (x-2)^n+1|
2/3*|1/(x-2)|<1
x belongs to (4/3, 8/3)
Therefore, the radius is 2/3

Use parentheses! What you wrote means
$$a_n= \frac{2n+1}{3n-1} a_n - 1$$
but I suspect you mean
$$a_n= \frac{2n+1}{3n-1} a_{n - 1}$$
In ASCII you should write a_n = [(2n+1)/(3n-1)] a_(n-1) or a(n)= [(2n+1)/(3n-1)] a(n-1)