Radius of convergence of the series

[yeahyeah<3]

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

Ratio Test for Absolute Convergence?

The Attempt at a Solution

Limit x--> infinity (2n+3) (x-2)^n+1 (3n-1) (x-2)^n
(3n+2) times (2n+1)

leaving the interval of convergence to be 1<x<3
which means the radius would be 1.
which isn't a choice.
What did I do wrong?
Thanks!

 

[HallsofIvy]

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

Ratio Test for Absolute Convergence?

The Attempt at a Solution

Limit x--> infinity (2n+3) (x-2)^n+1 (3n-1) (x-2)^n
(3n+2) times (2n+1)

leaving the interval of convergence to be 1<x<3
which means the radius would be 1.
which isn't a choice.
What did I do wrong?
Thanks!

With simple powers of n or, as in this case, linear terms in numerator and denominator, the radius of convergence is always 1.

 

[yeahyeah<3]

while that is true, that is not a choice.

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

 

[Dick]

while that is true, that is not a choice.

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

The ratio of the nth term to the (n-1)th term is a_n*(x-2)^n/(a_(n-1)*(x-2)^(n-1). The ratio of the two a's was given to you in the problem. What's the limit?

 

[yeahyeah<3]

Limit = (2/3) ?

However, according to the answer key the answer is (3/2).
I'm just looking for how to get that answer.

 

[Cyosis]

When the limit is 2/3, R=?

 

[Dick]

Limit = (2/3) ?

However, according to the answer key the answer is (3/2).
I'm just looking for how to get that answer.

Review the ratio test. How is the limiting ratio of the a's related to the radius of convergence?

 

[yeahyeah<3]

When the limit is 2/3, R=?

3/2 which makes sense but..
the limit is 1 when the R.T.A.C is applied?

 

[Cyosis]

I am not familiar with R.T.A.C, but $$\lim_{x \to +\infty} |\frac{a_n}{a_{n-1}}| =\lim_{x \to +\infty} \frac{2n+1}{3n-1} \neq 1$$

 

[yeahyeah<3]

Ratio Test for Absolute Convergence.
Basically to find the values of x for which the series converges.
when the test is applied the limit turns out to be one i believe... =/

 

[Dick]

Nooo. The ratio test says it converges at x if (2/3)*|x-2|<1. How does that turn into a radius of 1??

 

[Cyosis]

Well the limit I wrote down in the previous test is to my knowledge the ratio test. Could you perhaps write your R.T.A.C down?

 

[yeahyeah<3]

(2n+3) (x-2)^n+1 (3n-1) (x-2)^n
(3n+2) times (2n+1)

with a limit in front of that.. isn't that 6/6 = 1?

 

[Cyosis]

First of all you must not include the x term in your ratio test only $\frac{a_{n+1}}{a_n}$. Secondly I have no idea how you get 6/6 you take the limit of n goes to what exactly?

 

[yeahyeah<3]

n --> infinity
and my teacher has us include our x term
you sub in n+1 for all n
and then multiple by the 1/an

 

[Cyosis]

I think I see what you're doing now $$\lim_{x \to +\infty} \left|\frac{2n+3}{3n+2} \frac{(x-2)^{n+1}}{(x-2)^n} \right|$$. If this is indeed what you're doing show us how you got to 6/6 when n goes to infinity. If this is not what you're doing then write down the limit you are using. As it is now you keep me guessing most of the time, which isn't very effective.

 

[n!kofeyn]

I am not familiar with R.T.A.C, but $$\lim_{x \to +\infty} |\frac{a_n}{a_{n-1}}| =\lim_{x \to +\infty} \frac{2n+1}{3n-1} \neq 1$$

You have the ratio test wrong. The radius of convergence R is given by:
$$R = \lim_{n\to\infty} = \left| \frac{a_n}{a_{n+1}} \right|$$
if the limit exists.

This is what you need to apply to find the the radius of convergence of this series.

 

[Cyosis]

No I never claimed that my limit gave R, in fact I hinted that if the limit he finds is L then R=L^-1.

n --> infinity
and my teacher has us include our x term
you sub in n+1 for all n
and then multiple by the 1/an

From this I am pretty sure that he has been taught the ratio test they way I presented it so far, just that he includes the x^n as well which doesn't really matter. I see no reason to claim that this definition is wrong, unless you can show me why. Turning it up side down will most likely only add to his confusion.

 

[n!kofeyn]

No I never claimed that my limit gave R, in fact I hinted that if the limit he finds is L then R=L^-1.

That's fine, but you never stated that fact. I saw that you hinted if the limit is 2/3 then R is what? But that could mean the original poster may think R is just 2/3, as it seems as he is confused how to apply the ratio test in the first place. Turning it upside down shouldn't confuse as I stated explicitly that the limit gave you R, and properties of limits gives you the reciprocal of the limit as well.