Radius of Convergence Power Series

1. Apr 2, 2013

xspook

1. The problem statement, all variables and given/known data

Determine the radius of convergence and the interval of convergence og the folling power series:

n=0 to infinity
Ʃ=$\frac{(2x-3)^{n}}{ln(2n+3)}$

2. Relevant equations

Ratio Test

3. The attempt at a solution
Well I started with the ratio test but I have no clue where to go with it

$\frac{(2x-3)^{n+1}}{ln(2(n+1)+3)}$ * $\frac{ln(2n+3)}{(2x-3)^{n}}$

I know that I will be left with the 2x-3 and I can pull that out in front of the limit to use when doing the interval of convergence, but I am lost with the natural log. should I distribute the two and make it

$\frac{(2x-3)^{n+1}}{ln(2n+5)}$ * $\frac{ln(2n+3)}{(2x-3)^{n}}$

2x-3 Lim as n approaches infinity of : ln(2n+3)-ln(2n+5)??

Any help would be appreciated

2. Apr 2, 2013

Zondrina

The limit of $\frac{ln(2n+3)}{ln(2n+5)}$ as n goes to infinity is just 1.

Can you continue?

EDIT : It might be informative to wrap your limit in an absolute value.

Last edited: Apr 2, 2013
3. Apr 2, 2013

xspook

Thanks,

let me solve and see what I get

4. Apr 2, 2013

xspook

So I can now set up the interval of convergence to be

-1 < 2x-3 < 1
1 < x < 2

so any number between 1 and two will cause the series to converge

now i have to check endpoints seperately

x:1 Ʃ=$\frac{(2x-3)^{n}}{ln(2n+3)}$
$\frac{(-1)^{n}}{ln(2n+3)}$ which is an alternating series

the limit is 0 and it is also decreasing

X:2 Ʃ=$\frac{(2x-3)^{n}}{ln(2n+3)}$
$\frac{(1)^{n}}{ln(2n+3)}$

this limit is also 0

so the interval is [1,2]?

Radius of convergence would be $\frac{1}{2}$?

EDIT: when x=2 I feel like I have to do something different for solving for the limit? Or is it correct?

5. Apr 2, 2013

Zondrina

Your radius of convergence is what you get from the calculation of $lim_{n→∞} |\frac{a_{n+1}}{a_n}| = L$.

If L is finite, then your radius of convergence is 1/L.
If L is 0, then your radius of convergence is ∞.
If L = ∞, then your radius of convergence is 0.

Since you got 1 as the limit, the radius is R = 1/1 = 1. So your series will converge for |2x-3| < 1, aka 1 < x < 2. So you're doing good.

So now to get a concrete interval of convergence, you need to check the endpoints.

At x = 1, it looks like your series is conditionally convergent.

At x = 2, it appears your series diverges. Could you tell me why?

What's your interval now?

Last edited: Apr 2, 2013
6. Apr 2, 2013

xspook

At x = 2, I don't know why it is divergent. I thought it was convergent to zero. If it isnt convergent the interval would become [1,2)

7. Apr 2, 2013

Zondrina

Okay good about the interval. In the case of x=2, can you compare your series to anything? Think about minimizing the denominator.

8. Apr 2, 2013

xspook

Basically we have

$\frac{1}{ln(2n+3)}$

The limit of ln(2n+3) is ∞. $\frac{1}{∞}$ = 0

is there something I am missing?

9. Apr 2, 2013

Zondrina

You're missing the whole picture.

Why are you taking limits? All you have to do is check the convergence of your series at the endpoints.

So when you did x = 1. You saw $\sum |a_n|$ diverged, but $\sum (-1)^na_n$ converges by the alternating series test, so it was conditionally convergent.

Now when x = 2. You don't need the A.S.T. It's quite literally a simple comparison.

Last edited: Apr 2, 2013
10. Apr 2, 2013

xspook

How about:

If I use the comparison test for
$\frac{1}{ln(2n+3)}$ and compare with $\frac{1}{n}$

$\frac{1}{n}$ is a divergent p series
using the comparison test, $\frac{1}{ln(2n+3)}$ is also divergent???

11. Apr 2, 2013

Zondrina

Yes that's the idea.

12. Apr 2, 2013

xspook

Thank you Zondrina!

13. Apr 7, 2013

xspook

My professor gave me a 4 out of 10 possible points....he passed out the solutions to the problems and oddly enough I have the same answer.

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