Solving Limits: Find Interval & Radius of Convergence

[ironman]

Homework Statement

[/B]
I have to find the radius of convergence and convergence interval. So for what x's the series converge.
The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

Homework Equations

[/B]

(?)

The Attempt at a Solution

[/B]
I tried using

with an =

but i don't seem to get anywhere...
Is there another way of showing that the lim = 0 ?

 

[Mark44]

Homework Statement

[/B]
I have to find the radius of convergence and convergence interval. So for what x's the series converge.
The answer is supposed to be for every real number. So the interval is: (-∞, ∞).
So that must mean that the limit L = 0. So the radius of convergence [ which is given by R = 1/L ] is going to be ∞.

Homework Equations

[/B]
View attachment 76309View attachment 76306 (?)

You need to take the limit of the entire term in the summation, including the (x + 1)ⁿ factor.

The Attempt at a Solution

[/B]
I tried using View attachment 76307 with an = View attachment 76308

but i don't seem to get anywhere...
Is there another way of showing that the lim = 0 ?

Show us the work that you did in taking the limit of a_n+1/a_n. It's easy to get wrong with those factorials.

BTW, this is NOT a precalculus problem, so I moved it to the calculus section.

 

[ironman]

You need to take the limit of the entire term in the summation, including the (x + 1)ⁿ factor.

Show us the work that you did in taking the limit of a_n+1/a_n. It's easy to get wrong with those factorials.

BTW, this is NOT a precalculus problem, so I moved it to the calculus section.

Ah ok I'm sorry.
Yes indeed. But you would get:

So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

It's not possible to simplify this, right?

 

[Dick]

Ah ok I'm sorry.
Yes indeed. But you would get: View attachment 76323
So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

View attachment 76324 It's not possible to simplify this, right?

Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.

 

[Mark44]

Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.

Ah ok I'm sorry.
Yes indeed. But you would get: View attachment 76323

Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.

So you might as well just use the 'n-part', for the (x-c) is just a number (?)

I ended up with:

View attachment 76324 It's not possible to simplify this, right?

It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

 

[ironman]

Sure you can simplify it. Start with the (n+1)!/n! part. Remember what the definition of n! and (n+1)! are. A lot of terms cancel.

Ah of course!
(n+1)!/ n! = n+1.

I still don't know these rules. But I used :
n=3
4*3*2*1/3*2*1 = 4
= n+1

 

[ironman]

Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

Please don't post your work as images. When I quote what you wrote your stuff doesn't show up.Actually you get |x + 1|. It needs to be included in your limit calculations, especially for when you are finding the interval of convergence.
.
It can and should be simplified. (n + 1)! = (n + 1) * n! and (2n + 2)! = (2n + 2)(2n + 1)*(2n)!.

Hmm I see.
You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

Weird that my calculus book leaves the |x +c| out then...

 

[Dick]

Hmm I see.
You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)!
Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ?

Weird that my calculus book leaves the |x +c| out then...

You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.

 

[ironman]

You should keep simplifying until you have no more factorials. Take care of the (4n)!/(4n+4)! part as well.

I've no idea how to do that...

 

[Ray Vickson]

I've no idea how to do that...

Isn't it the case that $(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!$? Just keep using the definition of factorial.

 

[ironman]

Isn't it the case that $(4n+4)! = (4n+4)(4n+3)(4n+2)(4n+1) (4n)!$? Just keep using the definition of factorial.

Oh well... I blame it on my lack of sleep, ha.
So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)

 

[Dick]

Oh well... I blame it on my lack of sleep, ha.
So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)

Ok, so do you know how to find its limit?

 

[ironman]

Ok, so do you know how to find its limit?

Yes, I do. Thanks for the help all!