Radius of Convergence Power Series

[xspook]

Homework Statement

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

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

Homework Equations

Ratio Test

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

 

[STEMucator]

Homework Statement

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

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

Homework Equations

Ratio Test

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

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.

 

[xspook]

Thanks,

let me solve and see what I get

 

[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 separately

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?

 

[STEMucator]

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 separately

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?

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?

 

[xspook]

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

 

[STEMucator]

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

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

 

[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?

 

[STEMucator]

Basically we have

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

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

is there something I am missing?

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.

 

[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?

 

[STEMucator]

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?

Yes that's the idea.

 

[xspook]

Thank you Zondrina!

 

[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.